Triple cos squared

$\cos^2 x + \cos^22x+\cos^23x=1$

$\Rightarrow \cos^2x+(2\cos^2x-1)^2+(4\cos^3x-3\cos x)^2=1$

$\Rightarrow 1+6\cos^2 x-20\cos^4x+16\cos^6x=1$

Letting $\cos^2 x$ be $y$ , we will have

$16y^3-20y^2+6y=0$

$\Rightarrow y(2y-1)(4y-3)=0$

$\Rightarrow y = 0 , y = \frac{1}{2} ,y =\frac{3}{4}$

$\Rightarrow \cos x = 0, \cos x = \pm \frac{\sqrt{3}}{2},\cos x = \pm \frac{1}{\sqrt{2}}$

Considering the three cases,

Case 1: cos x = 0

We have x = $\frac{\pi }{2}$ and $\frac{3\pi }{2}$ as the solution for the given interval.

Case 2: $\cos x = \pm \frac{\sqrt{3}}{2}$

There are basically 4 solutions to this since we have to take into account all the 4 angles in the 4 quadrants. (The angles are $\frac{\pi}{6},\frac{5\pi}{6},\frac{7\pi}{6},\frac{11\pi}{6}$ )

Case 3: $\cos x =\pm \frac{1}{\sqrt{2}}$

The same reasoning as case 2. (The angles are $\frac{\pi}{4},\frac{3\pi}{4},\frac{5\pi}{4},\frac{7\pi}{4}$ )

Summing up the number of solutions, we will have 10.

Using the identity $cos (A+B)=cos A cos B - sin A sin B$ , we get $cos 2x = 2cos^2 x-1$ and $cos 3x= 4cos^3 x-3 cos x$

Letting $a=cos x$ , we get $\\ a^2+(2a^2-1)^2+(4a^3-3a^2)^2=1$

Simplifying this gives us $a^2(2a^2-1)(4a^2-3)=0$ . Thus, there are $5$ possible values of a, all of which has a magnitude less than $1$ . Hence, there are also $5$ possible values for $cos x$ . In the interval $[0,2 \pi ]$ , each value value of $cos x$ gives two possible values for $x$ . Therefore there are $10$ solutions for the given equation.

We can rewrite the initial equation to: $1+\cos 2x +1+\cos 4x +1+\cos 6x=2$

$3+ \cos 2x + 2\cos^2 2x -1 +4\cos^3 2x -3 \cos 2x=2$

$\cos 2x (2\cos 2x -1)(\cos 2x +1)=0$

This will lead to 3 cases:

Case 1: $\cos 2x=0$ , there are 4 roots in the interval [0, $2\pi$ ]: $\frac{\pi}{4},\frac{3\pi}{4},\frac{5\pi}{4},\frac{7\pi}{4}$ .

Case 2: $2 \cos 2x -1=0$ , there are 4 roots in the interval [0, $2\pi$ ]: $\frac{\pi}{6},\frac{11\pi}{6},\frac{5\pi}{6},\frac{7\pi}{6}$ .

Case 3: $\cos 2x +1=0$ , there are 2 roots in the interval [0, $2\pi$ ]: $\frac{\pi}{2},\frac{3\pi}{2}$ .

Overall, there are 10 roots in the interval [0, $2\pi$ ].

$\cos^2x + \cos^22x + \cos^23x = 1 \\ \implies 2\cos^2x + 2\cos^22x + 2\cos^23x = 2 \\ \implies \cos 2x + \cos 4x + \cos 6x = -1 \\ \implies 2\cos2x\cos4x + \cos4x = -1 \\ \implies \cos4x(2\cos2x + 1) = -1 \\ \implies (2cos^22x -1)(2cos2x+1) = -1 \\ \implies 2cos^32x+cos^2x - cos2x = 0 \\ \implies \cos2x(2\cos^22x+\cos2x -1) = 0 \\ \implies \cos2x(2\cos2x-1)(\cos2x+1) = 0 \\ \implies (\cos^2x-1)(4cos^2x-3)(cos^2x) = 0$ $$ This gives three cases which tell us $\cos x = \pm {\frac{\sqrt{3}}{4}}$ or $\cos x = 0$ or $\cos x = \pm1$ $$ The first case has four solutions, the second has two and the third has four again. So total solutions are $\boxed{4+2+4 = 10}$

$\cos^2 x + \cos^2 2x + \cos^2 3x =1$

$\Longrightarrow \frac {1+\cos 2x}{2} + \frac {1+\cos 4x}{2} + \frac {1+ \cos 6x}{2}=1$

$\Longrightarrow 1+\cos 2x + 1+ \cos 4x + 1+ \cos 6x =2$

$\Longrightarrow 3 + \cos 2x + 2 \cos^2 2x -1 + 4 \cos^3 2x - 3 \cos 2x =2$

$\Longrightarrow \cos 2x (2 \cos 2x-1)(\cos 2x+1)=0$

Now we have three cases...

Case 1 : $\cos 2x=0$ ... In the interval $[0,2\pi]$ , there are 4 possible zeroes... They are, $\frac {\pi}{4}, \frac {3\pi}{4}, \frac {5\pi}{4}, \frac {7\pi}{4}$

Case 2 : $2 \cos 2x-1=0$ ... In the interval $[0,2\pi]$ , there are 4 possible zeroes... They are, $\frac {\pi}{6}, \frac {5\pi}{6}, \frac {7\pi}{6}, \frac {11\pi}{6}$

Case 3 : $\cos 2x + 1=0$ ... In the interval $[0,2\pi]$ , there are 2 possible zeroes... They are, $\frac {\pi}{2}, \frac {3\pi}{2}$

Hence, there are total $4+4+2 = 10$ solutions in the interval $[0,2\pi]$

Therefore, the required answer is $\fbox {10}$

$\cos^2 x + \cos^2 2x + \cos^2 3x =1$

$(2\cos^2 x -1) + (2\cos^2 2x -1) + (2\cos^3 x -1) = -1$

$\cos 2x + \cos 4x + \cos 6x + \cos 0 = 0$

$2\cos 3x \cos x + 2\cos^2 3x = 0$

$2 \cos 3x (\cos x + \cos 3x) = 0$

$4 \cos x \cos 2x \cos 3x = 0$

$\cos x=0$ leads to the solutions $x = \dfrac{\pi}{2}$ and $x=\dfrac{3\pi}{2}$ .

$\cos 2x = 0$ leads to the solutions $x=\dfrac{\pi}{4}$ , $x=\dfrac{3\pi}{4}$ , $x=\dfrac{5\pi}{4}$ and $x=\dfrac{7\pi}{4}$ .

$\cos 3x = 0$ leads to the solutions $x=\dfrac{\pi}{6}$ , $x=\dfrac{3\pi}{6}$ , $x=\dfrac{5\pi}{6}$ , $x=\dfrac{7\pi}{6}$ , $x=\dfrac{9\pi}{6}$ and $x=\dfrac{11\pi}{6}$ .

However, some of these solutions are actually repeated, so ignoring these, we obtain $\boxed{10}$ solutions for the given expression.

Solve for x: cos^2(x)+cos^2(2 x)+cos^2(3 x) = 1 Rewrite the left hand side. Simplify trigonometric functions: 1+6 cos^2(x)-20 cos^4(x)+16 cos^6(x) = 1 Simplify 1+6 cos^2(x)-20 cos^4(x)+16 cos^6(x) = 1 by making a substitution. Simplify and substitute y = cos^2(x): 16 cos^6(x)-20 cos^4(x)+6 cos^2(x)+1 = 1+6 cos(x)^2-20 (cos(x)^2)^2+16 (cos(x)^2)^3 = 16 y^3-20 y^2+6 y+1 = 1: 16 y^3-20 y^2+6 y+1 = 1 Move everything to the left hand side. Subtract 1 from both sides: 16 y^3-20 y^2+6 y = 0 Factor the left hand side. The left hand side factors into a product with four terms: 2 y (2 y-1) (4 y-3) = 0 Divide both sides by a constant to simplify the equation. Divide both sides by 2: y (2 y-1) (4 y-3) = 0 Solve each term in the product separately. Split into three equations: y = 0 or 2 y-1 = 0 or 4 y-3 = 0 Look at the first equation: Perform back substitution on y = 0. Substitute back for y = cos^2(x): cos^2(x) = 0 or 2 y-1 = 0 or 4 y-3 = 0 Eliminate the exponent. Take the square root of both sides: cos(x) = 0 or 2 y-1 = 0 or 4 y-3 = 0 Solve for x. Take the inverse cosine of both sides: x = pi/2+pi n 1 for n 1 element Z or 2 y-1 = 0 or 4 y-3 = 0 Look at the second equation: Isolate terms with y to the left hand side. Add 1 to both sides: x = pi/2+pi n 1 for n 1 element Z or 2 y = 1 or 4 y-3 = 0 Solve for y. Divide both sides by 2: x = pi/2+pi n 1 for n 1 element Z or y = 1/2 or 4 y-3 = 0 Perform back substitution on y = 1/2. Substitute back for y = cos^2(x): x = pi/2+pi n 1 for n 1 element Z or cos^2(x) = 1/2 or 4 y-3 = 0 Eliminate the exponent on the left hand side. Take the square root of both sides: x = pi/2+pi n 1 for n 1 element Z or cos(x) = 1/sqrt(2) or cos(x) = -1/sqrt(2) or 4 y-3 = 0 Look at the second equation: Solve for x. Take the inverse cosine of both sides: x = pi/2+pi n 1 for n 1 element Z or x = pi/4+2 pi n 2 for n 2 element Z or x = (7 pi)/4+2 pi n 3 for n 3 element Z or cos(x) = -1/sqrt(2) or 4 y-3 = 0 Look at the fourth equation: Solve for x. Take the inverse cosine of both sides:

x = pi/2+pi n 1 for n 1 element Z

or x = pi/4+2 pi n 2 for n 2 element Z

or x = (7 pi)/4+2 pi n 3 for n 3 element Z

or x = (3 pi)/4+2 pi n 4 for n 4 element Z or x = (5 pi)/4+2 pi n 5 for n 5 element Z or 4 y-3 = 0

Look at the sixth equation: Isolate terms with y to the left hand side. Add 3 to both sides:

x = pi/2+pi n 1 for n 1 element Z

or x = pi/4+2 pi n 2 for n 2 element Z

or x = (7 pi)/4+2 pi n 3 for n 3 element Z

or x = (3 pi)/4+2 pi n 4 for n 4 element Z

or x = (5 pi)/4+2 pi n 5 for n 5 element Z or 4 y = 3

Solve for y. Divide both sides by 4:

x = pi/2+pi n 1 for n 1 element Z

or x = pi/4+2 pi n 2 for n 2 element Z

or x = (7 pi)/4+2 pi n 3 for n 3 element Z

or x = (3 pi)/4+2 pi n 4 for n 4 element Z

or x = (5 pi)/4+2 pi n 5 for n 5 element Z or y = 3/4 Perform back substitution on y = 3/4. Substitute back for y = cos^2(x):

x = pi/2+pi n 1 for n 1 element Z

or x = pi/4+2 pi n 2 for n 2 element Z

or x = (7 pi)/4+2 pi n 3 for n 3 element Z

or x = (3 pi)/4+2 pi n 4 for n 4 element Z

or x = (5 pi)/4+2 pi n 5 for n 5 element Z or cos^2(x) = 3/4

Eliminate the exponent on the left hand side. Take the square root of both sides:

x = pi/2+pi n 1 for n 1 element Z

or x = pi/4+2 pi n 2 for n 2 element Z

or x = (7 pi)/4+2 pi n 3 for n 3 element Z

or x = (3 pi)/4+2 pi n 4 for n 4 element Z

or x = (5 pi)/4+2 pi n 5 for n 5 element Z or cos(x) = sqrt(3)/2 or cos(x) = -sqrt(3)/2 Look at the sixth equation: Solve for x. Take the inverse cosine of both sides:

x = pi/2+pi n 1 for n 1 element Z

or x = pi/4+2 pi n 2 for n 2 element Z

or x = (7 pi)/4+2 pi n 3 for n 3 element Z

or x = (3 pi)/4+2 pi n 4 for n 4 element Z

or x = (5 pi)/4+2 pi n 5 for n 5 element Z or x = pi/6+2 pi n 6 for n 6 element Z or x = (11 pi)/6+2 pi n 7 for n 7 element Z or cos(x) = -sqrt(3)/2

Look at the eighth equation: Solve for x. Take the inverse cosine of both sides: Answer: |
| x = pi/2+pi n 1 for n 1 element Z

or x = pi/4+2 pi n 2 for n 2 element Z

or x = (7 pi)/4+2 pi n 3 for n 3 element Z

or x = (3 pi)/4+2 pi n 4 for n 4 element Z

or x = (5 pi)/4+2 pi n 5 for n 5 element Z or x = pi/6+2 pi n 6 for n 6 element Z

or x = (11 pi)/6+2 pi n 7 for n 7 element Z

or x = (5 pi)/6+2 pi n 8 for n 8 element Z or x = (7 pi)/6+2 pi n 9 for n 9 element Z

Finally we have 10 solutions.

We know that:

$cos(2x) = {cos}^2x - {sin}^2x = 2{cos}^2x - 1$ $cos(3x) = 4{cos}^3x - 3{cos}x$

So:

${cos}^2x + {cos}^22x + {cos}^23x =$ ${cos}^2x + (2{cos}^2x - 1)^2 + (4{cos}^3x - 3{cos}x)^2 = 1$ ${cos}^2x + 4{cos}^4x - 4{cos}^2x + 1 + {cos}^2x(4{cos}^2x - 3)^2 = 1$ ${cos}^2x(16{cos}^4x - 20{cos}^2x + 6) = 0$

Then:

${cos}^2x = 0$ or $16{cos}^4x - 20{cos}^2x + 6 = 0$

If ${cos}^2x = 0 \Rightarrow {cos}x = 0 \Rightarrow x = 90^\circ$ or $x = 270 ^\circ$

Else if $16{cos}^4x - 20{cos}^2x + 6 = 0 \Rightarrow {cos}^2x = \frac 12$ or ${cos}^2x = \frac 34$

If ${cos}^2x = \frac 12 \Rightarrow x = \pm \frac {\sqrt{2}}2$ , x can be $45^\circ, 135 ^\circ, 225 ^\circ, 315 ^\circ$

If ${cos}^2x = \frac 34 \Rightarrow x = \pm \frac {\sqrt{3}}2$ , x can be $30^\circ, 150 ^\circ, 210 ^\circ, 330 ^\circ$

So, there are 10 solutions.

Using the fact that $\cos2x=2\cos^2x-1$ and $\cos3x=4\cos^3x-3\cos x,$ we can substitute into our equation: $\begin{aligned} \cos^2x+\cos^22x+\cos^23x&=&1\\ \cos^2x+(2\cos^2x-1)^2+(4\cos^3x-3\cos x)^2&=&1. \end{aligned}$ Expanding gives $\begin{aligned} \cos^2x+4\cos^4x-4\cos^2x+1+16\cos^6x-24\cos^4x+9\cos^2x&=&1\\ 16\cos^6x-20\cos^4x+6\cos^2x&=&0. \end{aligned}$ Dividing both sides by $2$ and factoring yields $\begin{aligned} 16\cos^6x-20\cos^4x+6\cos^2x&=&0\\ \cos^2x(2\cos^2x-1)(4\cos^2x-3)&=&0. \end{aligned}$ From here we get that $\begin{array}{ccc|ccc|ccc} \cos^2x&=&0&\cos^2x&=&\tfrac12&\cos^2x&=&\tfrac34\\ \cos x&=&0&\cos x&=&\pm\tfrac{\sqrt2}{2}&\cos x&=&\pm\tfrac{\sqrt3}{2}\\ x&=&\tfrac\pi2,\tfrac{3\pi}2&x&=&\tfrac\pi4,\tfrac{3\pi}4,&x&=&\tfrac\pi6,\tfrac{5\pi}6\\ &&&&&\tfrac{5\pi}4,\tfrac{7\pi}4&&&\tfrac{7\pi}6,\tfrac{11\pi}6 \end{array}$ There are a total of $\boxed{10}$ solutions.

every 180 degree there are 5 solution 30,45,90,135,150 degree

Let z=cos x, z^2+(2z^2-1)^2+(4z^3-3z)^2=1, then we can simplify this equation to: 8z^6-10z^4+3z^2=0, that can be factorized to z^2(4z^2-3)(2z^2-1)=0. The solutions are: z=0, or x= π÷2, 3π÷2, and z=(3/4)^0.5, or x=30°, 330°, and z =-(3/4)^0.5, or x=150°, 210°, and z= +/ - (1/2)^0.5, or x = (45°,315°,135°,225°). So there are in total 10 solutions

cos^2 x+cos^2 2x+cos^2 3x=1

(1/2)+(1/2)cos2x+(1/2)+(1/2)cos4x(1/2)+(1/2)cos6x=1

cos4x(cos2x+1/2)-1/2=0

(2cos^2 2x -1)(cos2x+1/2)-1/2=0

cos2x(cos^2 2x+cos2x-1)=0

cos2x=0 have 4 solution.

cos2x=1/2 have 4 solution

cos 2x=-1 have 2 solution.

all solutions are 10.

With $\cos ax=\frac{e^{axi}+e^{-axi}}{2}$ the problem becomes $\displaystyle \frac 1 4 \left( e^{2*xi}+2+e^{-2*xi} + e^{2*2xi}+2+e^{-2*2xi} + e^{2*3xi}+2+e^{-2*3xi} \right) = 1$ or $\displaystyle e^{6xi}+e^{4xi}+e^{2xi}+e^{-2xi}+e^{-4xi}+e^{-6xi} = 4-6=-2$ If we add 1 to the LHS (and the RHS), the LHS becomes a geometric series, so multiplying by $\sin x=\frac{e^{ix}-e^{-ix}}{2i}$ , we get $\displaystyle \frac {e^{7xi}-e^{-7xi}}{2i} = -\sin x$ and finally $\sin(7x)=\sin(-x)$ . Since we multiplied by $\sin x$ we may have introduced solutions, so the solutions where $\sin x=0$ are candidate solutions that we have to check. So we are left with candidate solutions

$\displaystyle 7x=-x+2k\pi, k \in {\mathbb Z}$ or $\displaystyle 7x=\pi+x+2k\pi, k \in {\mathbb Z}$

Rewritten:

$\displaystyle x=\frac k 4 \pi, k \in {\mathbb Z}$ or $\displaystyle x=\frac \pi 6 +\frac k 3 \pi, k \in {\mathbb Z}$ .

The candidate solutions $x=k\pi$ aren't solutions to the original question. With taking the domain into considering this leads to the answer: there are $\boxed {10}$ solutions: $\frac 16\pi, \frac 14\pi, \frac 12\pi, \frac 34\pi, \frac 56\pi, \frac 76\pi, \frac 54\pi, \frac 32\pi, \frac 34\pi \text{ and } \frac {11}6\pi$

$cos(2x) = 2 \, cos^{2}(x) - 1$

$cos(3x) = [ e^{i \cdot 3x} + e^{-i \cdot 3x} ]/2 = [(e^{ix})^{3} + (e^{-ix})^{3} ]/2$

Using Euler's Formula the above becomes

$cos(3x) = cos^{3}(x) - 3 \, cos(x) \, sin^{2}(x)$

These give us

$cos^{2}(x) + cos^{2}(2x) + cos^{2}(3x) = 1 - 3 \, cos^{2}(x) + 4 \, cos^{4}(x) - 6 \, cos^{4}(x) \, sin^{2}(x) + 9 \, cos^{2}(x) \, sin^{4}(x) + cos^{6}(x) = 1$

$0 = -3 \, cos^{2}(x) + 4 \, cos^{4}(x) - 6 \, cos^{4}(x) \, sin^{2}(x) + 9 \, cos^{2}(x) \, sin^{4}(x) + cos^{6}(x)$

A solution to this is $cos(x) = 0$ . On the interval $[0, 2\pi ]$ this is $\large x=(\pi /2, \; 3\pi /2)$ . If $cos(x) \neq 0$ then you can divide by $cos^{2}(x)$

$0 = -3 + 4 \, cos^{2}(x) - 6 \, cos^{2}(x) \, sin^{2}(x) + 9 \, sin^{4}(x) + cos^{4}(x)$

Substituting $sin^{2}(x) = 1 - cos^{2}(x)$ into the above equation for $sin^{2}(x)$ it reduces to

$0 = 16\, cos^{4}(x) - 20 \, cos^{2}(x) + 6 \; \; \; \; \; \; \;$ Using the Quadratic Formula gives

$cos^{2}(x) = (3/4, \; 1/2) \; \; \; \; \;$ or

$cos(x) = ( \frac{1}{2} \sqrt{3}, \; \sqrt{ \frac{1}{2} }, \; - \frac{1}{2} \sqrt{3}, \; - \sqrt{ \frac{1}{2} } )$

On the interval $[0, \; 2 \pi ]$ each of these four cosine values yield two values of $x$ . These eight along with the two solutions above make a total of ten solutions. The full solution set is

$\large x = (\pi /2, \; 3\pi /2, \; \pi /6, \; \pi /4, \; 5 \pi /6, \; 3 \pi /4, \; 11 \pi /6, \; 7 \pi /4, \; 7 \pi /6, \; 5 \pi /4)$

To begin with, let us note that

• $\cos2\alpha=\cos^{2}\alpha - \sin^{2}\alpha$
• $\cos(\alpha+\beta) = \cos \alpha \cos \beta - \sin \alpha \sin \beta$
• $\sin2a =2 \sin \alpha \cos \alpha$

Therefore, if we add $1$ to both sides of the expression we would get

$\cos2x+1=1+\cos^{2}x-\sin^{2}x$

Taking into account that $\sin^{2}x+\cos^{2}x=1 \Rightarrow 1-\sin^{2}x=\cos^{2}x$ , we may note that

$1+\cos^{2}2x=2cos^{2}x \Rightarrow \frac{1+cos^{2}2x}{2}=cos^{2}x$

Similarly, if instead of $x$ we replace $2x$ or $3x$ in the expansion, we may derive that

$\cos^{2}2x=\frac{1+cos4x}{2}$ and $\cos^{2}3x=\frac{1+cos6x}{2}$ .

Therefore, the initial equation would take the form of

$\frac{1+\cos2x+1+\cos4x+1+\cos6x}{2}=1 \Rightarrow$

$\Rightarrow 1+\cos2x+1+\cos4x+1+\cos6x=2$

Furthermore, let us see that

$\cos4x=\cos^{2}2x-\sin^{2}2x=\cos^{2}2x-(1-\cos^{2}2x)=$

$=2\cos^{2}2x-1$

and consequently

$\cos6x=\cos(2x+4x)=\cos4x\cos2x-\sin4x\sin2x=$

$=(2cos^{2}2x-1)\cos2x-2\sin2x\cos2x\sin2x=$

$=2\cos^{3}2x-\cos2x-2\sin^{2}2x\cos2x=$

$=2\cos^{3}2x-\cos2x-2(1-\cos^{2}2x)\cos2x=$

$=2\cos^{3}2x-\cos2x-2\cos2x+2\cos^{3}2x=$

$=4\cos^{3}2x-3\cos2x$

Thus, we finally get the equation to the form of

$3+\cos2x+2\cos^{2}2x-1+4\cos^{3}2x-3\cos2x=2 \Rightarrow$

$\Rightarrow 4\cos^{3}2x+2\cos^{2}2x-2\cos2x=0 \Rightarrow$

$\Rightarrow 2\cos^{3}2x+\cos^{2}2x-\cos2x=0 \Rightarrow$

$\Rightarrow \cos2x(2\cos^{2}2x-\cos2x-1)=0$

Since the only way to get $0$ from any integer is to multiply it with exactly $0$ , it is evident that either $\cos2x=0$ or $2\cos^{2}2x+\cos2x-1=0$ .

However, before we jump to the separate cases, let us have a look at the equation $2\cos^{2}2x+\cos2x-1=0$ .

If we were to substitute $\cos2x$ with $m$ (making sure that $-1\leq m\leq1$ ), we would get $2m^{2}+m-1=0$ . The roots to this quadratic equation are $-1$ and $\frac{1}{2}$ , which means that if factorized, the equation would take the form of $2(x+1)(x-\frac{1}{2}=(x+1)(2x-1)$ . This is due to the fact that any equation of the form $a_{n}x^n+a_{n-1}x^{n-1}+\ldots +a_{2}x^{2}+a_{1}x+a_{0}=0$ can be expressed as $a_{n}(x-x_{n})(x-x_{n-1})\ldots (x-x_{2})(x-x_{1}) =0$ , where $x_{n},x_{n-1},\ldots x_{2},x_{1}$ are the roots of the given equation.

Regardless, by substituting back $x$ with $\cos2x$ , we get $2\cos^{2}2x+\cos2x-1=(\cos2x+1)(2\cos2x-1)$ , which means that

$\cos2x(\cos2x+1)(2\cos2x-1)=0$

For this to be true, we must consider all $3$ cases:

Case 1:

$\cos2x=0 \Rightarrow x=45^{\circ}+k\pi$ , where $k$ is an integer.

This case would be true in the interval of $[0,2\pi]$ for $4$ different values for $x$ , which are $x_{1}=45^{\circ}=\frac{\pi}{4}, x_{2}=135^{\circ}=\frac{3\pi}{4},x_{3}=225^{\circ}=\frac{5\pi}{4}$ and $x_{4}=315^{\circ}=\frac{7\pi}{4}$ .

Case 2:

$2\cos2x-1=0 \Rightarrow \cos2x=\frac{1}{2}$

Since in the original equation all of the $\cos x,\cos2x$ and $\cos3x$ have all been squared, we may include the solutions where $\cos2x=-\frac{1}{2}$ as well, since both solutions would yield the same result. Thus, in the interval $[0,2\pi]$ we are left with $4$ more solutions which are $x_{5}=30^{\circ}=\frac{\pi}{6},x_{6}=150^{\circ}=\frac{5\pi}{6},x_{7}=210^{\circ}=\frac{7\pi}{6}$ and $x_{8}=330^{\circ}=\frac{11\pi}{6}$ .

Case 3:

$\cos2x+1=0 \Rightarrow \cos2x=-1$ .

Again, since in the original equation all of the $\cos x,\cos2x$ and $\cos3x$ have all been squared, we may include the solution for $\cos2x=1$ as well, since both of the solutions would be yield the same result. Thus, in the interval $[0,2\pi]$ we get $2$ more solutions which are $x_{9}=90^{\circ}=\frac{\pi}{2}$ and $x_{10}=270^{\circ}=\frac{3\pi}{2}$ .

Furthermore, from all cases we get $4+4+2=10$ possible solutions, meaning that $\cos^{2}x+\cos^{2}2x+\cos^{2}3x=1$ has $10$ possible solutions for $x$ in the interval $[0,2\pi]$ .

Therefore, the answer is $10$ .

cos^2 X + cos^2 2X + cos^2 3X = 1 (1)

Because : sin^2 X + cos^2 X = 1

we get,

sin^2 X + sin^2 2X + sin^2 3X = 2 (2)

from Equation (1) and (2): (2) - (1):

(cos^2 X - sin^2 X) + (cos^2 X - sin^2 X) + (cos^2 X - sin^2 X) = -1

cos 2X + cos 4X + cos 6X = -1 (cos 2X + cos 6X) + cos 4X = -1

2 cos 4X cos 2X + cos 4X = -1

cos 4X ( 2 cos 2X + 1) = -1 (2 cos^2 2X - 1) (2 cos 2X + 1) = -1

4 cos^3 2X + 2 cos^2 2X - 2 cos^2 2X -1 = -1 4 cos^3 2X + 2 cos^2 2X - 2 cos^2 2X = 0

2 cos 2X (2 cos 2X - 1) (cos 2X + 1) = 0

So,

cos 2X = 0 or cos 2X = 1/2 or cos 2x = -1

cos 2X = 0, The Solution Set = {1/4 pi, 3/4 pi, 5/4 pi, 7/4 pi}

cos 2X = 1/2, The Solution Set = {1/6 pi, 5/6 pi, 7/6 pi, 11/12 pi}

cos 2X = -1, the Solution Set = {1/2 pi, 3/2 pi}

The Answer = 10

cos^2x+cos^2(2x)+cos^2(3x)=1 =>

(1+cos2x)/2+(1+cos4x)/2+(1+cos6x)/2=1 => 1+cos2x+1+cos4x+1+cos6x=2 => 3+cos2x+2cos^2(2x)−1+4cos^3(2x)−3cos2x=2 => cos2x(2cos2x−1)(cos2x+1)=0

now each factor will equal to zero so we have three cases

Case 1: cos2x=0, then in the interval [0,2π]: π/4,3π/4,5π/4,7π/4.

Case 2: 2cos2x−1=0, then in the interval [0,2π]: π/6,11π/6,5π/6,7π/6.

Case 3: cos2x+1=0, then n the interval [0,2π]: π/2,3π/2.

so total 4+4+2=10 roots in the interval [0,2π].

$cos^{2}x + cos^{2}2x + cos^{2}3x = 1$

$cos^{2}x + (2cos^{2}x - 1)^{2} + (4cos^{3}x - 3cosx)^{2} = 1$

$cos^{2}x + 4cos^{4} - 4cos^{2}x + 1 + 16cos^{6}x - 24cos^{4}x + 9cos^{2}x = 1$

$16cos^{6}x - 20cos^{4}x + 6cos^{2}x = 0$

Case 1

For $cos^{2}x = 0$

$x = \frac{\pi}{2} , \frac{3\pi}{2}$

Case 2

For $cos^{2}x ≠ 0$

$16cos^{4}x - 20cos^{2}x + 6 = 0$

$(2cos^{2}x -1)(4cos^{2}x - 3)= 0$

hence , $cosx = \frac{\pi}{6} , \frac{5\pi}{6} , \frac{7\pi}{6} , \frac{11\pi}{6} , \frac{\pi}{4} , \frac{3\pi}{4}, \frac{5\pi}{4} , \frac{7\pi}{4}$

hence number of solutions are 10

Do trigonometric addition:

$x \cos ^2+(2 x) \cos ^2+(3 x) \cos ^2=1\\ 16 x \cos ^6-20 \left(x \cos ^4\right)+6 x \cos ^2$

Substitute $y=\cos ^2(x)$ :

$16 y^3-20 y^2+6 y=0\\ y=\left\{0,\frac{1}{2},\frac{3}{4}\right\}\\ \cos ^2(x)=\left\{0,\frac{1}{2},\frac{3}{4}\right\}$

We go case by case and find solutions within the interval $[0,2 \pi ]$ .

Case 1: $\cos ^2(x)=0$

$\cos ^2(x)=0\\ \cos (x)=0\\ x=2 \pi c_1\pm \frac{\pi }{2}\\ x=\left\{\frac{\pi }{2},\frac{3 \pi }{2}\right\}$

Case 2: $\cos ^2(x)=\frac{1}{2}$

$\cos ^2(x)=\frac{1}{2}\\ \cos (x)=\pm\frac{1}{\sqrt{2}}\\ x=\left\{2 \pi c_2\pm \frac{\pi }{4},2 \pi c_3\pm \frac{3 \pi }{4}\right\}\\ x=\left\{\frac{\pi }{4},\frac{3 \pi }{4},\frac{5 \pi }{4},\frac{7 \pi }{4}\right\}$

Case 3: $\cos ^2(x)=\frac{3}{4}$

$\cos ^2(x)=\frac{3}{4}\\ \cos (x)=\pm\frac{\sqrt{3}}{2}\\ x=\left\{2 \pi c_4\pm \frac{\pi }{6},2 \pi c_5\pm \frac{5 \pi }{6}\right\}\\ x=\left\{\frac{\pi }{6},\frac{5 \pi }{6},\frac{7 \pi }{6},\frac{11 \pi }{6}\right\}$

Hence there are a total of 10 distinct solutions.