Ren's trigonometric equation

The equation can be arranged as:

$\sin^4\theta+\sin^2\theta\cos^2\theta+\cos^4\theta=\frac{3}{4}$

$\sin^2\theta(\sin^2\theta+\cos^2\theta)+\cos^4\theta=\frac{3}{4}$

Since $\sin^2\theta+\cos^2\theta=1$ ,

$\sin^2\theta+\cos^4\theta=\frac{3}{4}$

$1-\cos^2\theta+\cos^4\theta=\frac{3}{4}$

$\cos^2\theta-\cos^4\theta=\frac{1}{4}$

$\cos^2\theta(1-\cos^2\theta)=\frac{1}{4}$

$\cos^2\theta\sin^2\theta=\frac{1}{4}$

$\cos\theta\sin\theta=\pm\frac{1}{2}$

$2\cos\theta\sin\theta=\pm1$

$\sin2\theta=\pm1$

When $\sin2\theta=1$ ,

$2\theta={90^\circ,450^\circ\ldots}$

$\theta={45^\circ,225^\circ}$ Note that: $0^\circ\leq\theta\leq360^\circ$

When $\sin2\theta=-1$ ,

$2\theta={270^\circ,630^\circ\ldots}$

$\theta={135^\circ,315^\circ}$ Note that: $0^\circ\leq\theta\leq360^\circ$

$\therefore\theta={45^\circ,135^\circ,225^\circ,315^\circ}$

The sum of all possible values of $\theta=45^\circ+135^\circ+225^\circ+315^\circ=720$

First we will convert the expression to make it a complete square. Hence $\sin^4 \theta + \cos^4 \theta + 2 \sin^2 \theta \cos^2 \theta - \sin^2 \theta \cos^2 \theta = \frac{3}{4}$ $(\sin^2 \theta + \cos^2 \theta)^2 - \sin^2 \theta \cos^2 \theta = \frac{3}{4}$

But we already know that $\sin^2 \theta + \cos^2 \theta = 1$

Hence the expression becomes > $1 - \sin^2 \theta \cos^2 \theta = \frac{3}{4}$ Now we need to use the multiple angle formula that is $2\sin\theta\cos\theta = \sin 2\theta$ So, $4\sin^2\theta \cos^2\theta = \sin^2 2\theta \Rightarrow \sin^2\theta \cos^2\theta = \frac{\sin^2 2\theta}{4}$ Hence the expression becomes > $1 - \frac{\sin^2 2\theta}{4} = \frac{3}{4}$ This gives > $\sin 2\theta = \pm 1$

Since $0^\circ \leq \theta \leq 360^\circ \Rightarrow 0^\circ \leq 2\theta \leq 720^\circ$ , we can state that >

$2\theta = 90^\circ, 270^\circ, 450^\circ, 630^\circ$ $\theta = 45^\circ, 135^\circ, 225^\circ, 315^\circ$

SO sum of all possible values of $\theta = 45 + 135 +225 +315 = 720$

Obviously $\sin^4\theta+\sin^2\theta\cos^2\theta+\cos\theta^4=1-\frac{1}{4}\sin^2(2\theta)$ . The equation $\sin(2\theta)=1$ has solutions $45, 135, 225, 315$ (in degrees), hence the result is $720$ .

Adding $\sin^2\theta \cos^2\theta$ to the equation we get $1=\dfrac{3}{4}+ \sin^2\theta \cos^2\theta$ which means $\sin2\theta = 1$ or $\sin2\theta = -1$ . Easily we get $\theta=45^o,135^o,225^o,315^o$ . So the sum of all possible value of $\theta$ is $45^o+135^o+225^o+315^o=720^o$ .

$\sin^{4}θ + \sin^{2}\cos^{2} +\cos^{4}θ$ $= (\sin^{2}θ + \cos^{2}θ)^{2} - \sin^{2}θ\cos^{2}θ$

$= 1 - \frac{\sin^{2}2θ}{4}$ $= \frac{3}{4}$

$\sin^{2}2θ = 1$

$\frac{1 - \cos4θ}{2} = 1$

$\cos4θ = -1$

so $θ = \frac{1}{4} (2nπ - π)$ or $\frac{1}{4} (2nπ + π)$

so $θ = \frac{π}{4} ,\frac{3π}{4} , \frac{5π}{4} , \frac{7π}{4}$

Answer is $\frac{π}{4} + \frac{3π}{4} + \frac{5π}{4} + \frac{7π}{4}$ = 4π = 720

$\sin^{4}\theta + \sin^{2}\theta \cos^{2}\theta + \cos^{4}\theta = (\sin^{2}\theta + \cos^{2}\theta)^{2} - \sin^{2}\theta \cos^{2}\theta = 1 - \sin^{2}\theta \cos^{2}\theta = 3/4$ $\sin^{2}\theta \cos^{2}\theta = 1/4$ $\sin\theta \cos\theta = 1/2, -1/2$ $\sin(2\theta) = 1, -1$ $2\theta = 90, 270, 450, 630$ $\theta = 45, 135, 225, 315$ $45 + 135 + 225 + 315 = 720$

simplify the expression to sin(theta)*cos(theta)=+-1/2 using sin^2 (theta)+cos^2 (theta)=1

(Sin^2 x+ cos^2 x)^2 - sin^2 x cos^2 x=3/4 =>1- sin^2 x cos^2 x=3/4 1-1/4(2sin x cos x)^2=3/4 1-1/4[sin^2(2x)]=3/4 4- sin^2(2x)=3 sin^2(2x)=1 sin2x=+_1 2x=90 or 270 or 450 and 630 or x=45 or 135 or 225 or 315. => x = 45+135+225+315=720

The given equation can be rewritten as $(\sin^{2}\theta+ \cos^{2}\theta)^{2} -\sin^{2}\theta\cos^{2}\theta=\frac{3}{4} \Rightarrow1-\sin^{2}\theta\cos^{2}\theta=\frac{3}{4} \Rightarrow\sin^{2}\theta\cos^{2}\theta=\frac{1}{4} \Rightarrow\sin\theta\cos\theta=\frac{1}{2} or \frac{-1}{2} \Rightarrow2\sin\theta\cos\theta=1 or -1 \Rightarrow\sin{2}\theta=1 \Rightarrow\theta=45 or 225 \Rightarrow\sin{2}\theta=-1 \Rightarrow\theta=135 or 315 \Rightarrow sum of all values of \theta=720$

Staff test solution.

The equation can be written as:

$\sin \theta$ ^4+2( $\sin \theta$ $\cos \theta$ )^2+ $\cos \theta$ ^4-( $\sin \theta$ $\cos \theta$ )^2= $\frac{3}{4}$

(sin^theta+cos^2theta)^2-( $\sin \theta$ $\cos \theta$ )^2= $\frac{3}{4}$

1-( $\sin \theta$ $\cos \theta$ )^2= $\frac{3}{4}$

( $\sin \theta$ $\cos \theta$ )^2= $\frac{1}{4}$

$\sin \theta$ $\cos \theta$ =+- $\frac{1}{2}$

$\sin \2theta$ \=+-1

The graph of sin2theta will give solutions 45, 135,225,315

The following expression can be written as $(sin ^2\theta + cos^2\theta )^2$ = $\frac{3}{4}$ $+ \sin^2\theta + cos^2 \theta$

= $1^2 = \frac {3}{4} + \sin^2 \theta + \cos^2 \theta$

implies \(\frac{1}{4} = \sin^2\theta + \cos^2\theta)\

On multiplying both sides by 4, 1 =\( 4\sin^2\theta +\cos^2\theta=sin^2 2\theta\)

implies $\sin 2\theta$ = $+1 or -1$ hence there are 4 values of theta within the limits that satisfy the following $\theta = 45, 135 , 225, 315$

therefore answer = $45 + 135+225+315= 720$

$\sin ^{4} \theta + \sin ^{2} \theta \cos ^{2} \theta + \cos ^{4} \theta = \frac{3}{4}$

Rewrite as

$(\sin ^{2} \theta + \cos ^{2} \theta) ^{2} - \sin ^{2} \theta \cos ^{2} \theta = \frac{3}{4}$

Use identity: $\sin ^{2} \theta + \cos ^{2} \theta = 1$

$- \sin ^{2} \theta \cos ^{2} \theta = - \frac{1}{4}$

$4 \sin ^{2} \theta \cos ^{2} \theta = 1$

$(2 \sin \theta \cos \theta) ^ {2} = 1$

$\sqrt{(2 \sin \theta \cos \theta) ^ {2}} = +/- 1$

$2 \sin \theta \cos \theta = +/- 1$

$\sin (2 \theta) = +/- 1$

$\theta = 45, 135, 225, 315$

$45 + 135 + 225 + 315 = 720$

We can write this way : 3/4 = 1 - [sin(θ).cos(θ)]^2. So sin(θ).cos(θ) = +-1/2. Using the fundamental law of the trigonometry we will have a equation ( 4[cos(θ)]^4 - 4[cos(θ)]^2 +1=0). Lastly we will have like solution : S = {45,135,225,315}, <graus>. The sum is 720.

sin^4 θ+2sin^2 θcos^2 θ+cos^4 θ=3/4+sin^2 θcos^2 θ

(sin^2 θ+cos^2 θ)^2=3/4+sin^2 θcos^2 θ

1=3/4+sin^2 θcos^2 θ

1/4=sin^2 θcos^2 θ

1/4=1/4(sin2θ)^2

(sin2θ)=+-1

(sin2θ)=1 then θ=45,225

sin 2θ=-1 then θ=135,315

so the sum of all θ are 45+225+135+315=720

Let $x=\sin\theta$ and $y=\cos\theta$ . We rewrite the given expression as: $$(x^2+y^2)^2=\frac{3}{4}+x^2y^2$$ Since $x=\sin\theta$ and $y=\cos\theta$ , the expression on the left is equivalent to $1$ .

We now have:

$$1=\frac{3}{4}+x^2y^2$$ $$\frac{1}{4}=x^2y^2$$ $$\frac{1}{2}=xy$$ In order for this to be true, $x$ and $y$ must have values of $\frac{\sqrt{2}}{2}$ and $-\frac{\sqrt{2}}{2}$ , in some permutation.

The values of $\theta$ corresponding to all possible values of $x$ and $y$ that are solutions are thus: $45^\circ, 135^\circ, 225^\circ,$ and $315^\circ$ .

The sum of these values is: $$S=45+135+225+315$$ $$S=\boxed{720}$$

Using that $­ \sin^2 \theta = 1 - \cos^2 \theta$ we get an equation with just cosine: $\cos^4 \theta - \cos^2 \theta + \frac{1}{4} = 0$ which gives $\cos^2 \theta = \frac{1}{2}$ and $\cos \theta = \pm \sqrt{\frac{1}{2}} = \pm \frac{\sqrt{2}}{2}$ . We know from the unit circle that $\cos \theta = \frac{\sqrt{2}}{2}$ has the solution $\theta = 45^\circ$ in the first quadrant, so by symmetry of the unit circle we obtain $\theta \in \{ 45^\circ , 135^\circ , 225^\circ , 315^\circ \}$ . The sum of these angles is 720 and is thus the answer to the problem.

Sin^4(theta) + 2.sin^2(theta).cos^2(theta) + cos^4(theta) - sin^2(theta).cos^2(theta) = 1 - sin^2(theta).cos^2(theta) =3/4 <=> 4sin^2(theta).cos^2(theta) = 1 <=> sin(2theta) = +-1 <=> theta = 45, 135, 225, 315. So we get the sum equal 45+135+225+315 = 720

$sin^2\theta+cos^2\theta = 1$

So $\left(sin^2\theta+cos^2\theta\right)^2 = sin^4\theta + 2sin^2\theta cos^2\theta + cos^4\theta = 1$

Subtracting the original equation we get $sin^2\theta cos^2\theta = \frac{1}{4}$

So $sin\theta cos\theta = \frac{1}{2}$

$cos\theta = \sqrt{1-sin^2\theta}$

$sin^2\theta \left( 1-sin^2\theta\right) = \frac{1}{4}$

$sin^4\theta -sin^2\theta + \frac{1}{4} = 0$

$\left(sin^2\theta +\frac{1}{2}\right)\left(sin^2\theta - \frac{1}{2}\right) = 0$

So $sin\theta = \pm \frac{1}{\sqrt{2}}$

Therefore $\theta = 45^{\circ}, 135^{\circ}, 225^{\circ}, 315^{\circ}$

$45+135+225+315 = 720$