Trigonometry Power Seven?

$\begin{aligned} \cos^7 x + \sin^4 x & = & 1 \\ \cos^7x & = & 1 - \sin^4 x \\ & = & (1 - \sin^2 x)(1 + \sin^2 x ) \\ & = & \cos^2 x (1 + \sin^2 x), \text{ now divide by } \cos^2 x \\ \cos^5 x & = & 1 + \sin^2 x \\ \cos^5 x+ \cos^2 x & = & 2 \\ \end{aligned}$

Since we have perform the operation of division of $\cos^2 x$ , then $\cos x = 0$ is a solution.

Furthermore, because sum of two functions with maximum value of $1$ gives a value of $2$ . Then $\cos^5 x = \cos^2 x = 1 \Rightarrow \cos x = 1$

So $\cos x = 0, 1 \Rightarrow x = \pm \frac \pi 2, 0$ , thus there's $\boxed{3}$ solution.

Rewrite the equation as

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

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

$\Longrightarrow (1 + \cos^{3}(x))\cos^{4}(x) = 2\cos^{2}(x).$

So either $\cos(x) = 0$ , which on $(-\pi, \pi)$ occurs at $x = -\dfrac{\pi}{2}, \dfrac{\pi}{2}$ , or

$(1 + \cos^{3}(x))\cos^{2}(x) = 2.$

Now given that $-1 \le \cos(x) \le 1$ we see that this equation can only be satisfied when $\cos(x) = 1$ , which on $(-\pi, \pi)$ occurs at $x = 0.$

Thus on the given interval there are $\boxed{3}$ solutions.

Simplest solution: cos⁷x + sin⁴x < cos²x + sin²x = 1 unless either cos x = 1 or sin x = 1 or -1. Hence, x = 0, pi/2 or -pi/2 are the only solutions.

Simple solution: have 90 (both plus and minus) and 0. One of the functions will turn to 1 if the angle is quadrantal, even at negative point.

cos^7(x)+sin^4(x)=1 <=>cos^7(x)+(1-cos^2(x))²=1 <=>cos^7(x)+cos^4(x)-2cos^2(x)+1=1 <=>cos^7(x)+cos^4(x)-2cos^2(x)=0 <=>cos^2(x)(cos^5(x)+cos^2(x)-2)=0 <=>cos^2(x)=0 or cos^5(x)+cos^2(x)-2=0 <=>cosx=0 -π≤x≤π =>x=±½π and x=0 => This equation has 3 roots.