find all solutions of the equation in the interval 0, 2π).\n4 cos x = -sin²x + 1\nwrite your answer in…

Explanation:

Step1: Use trig identity $\sin^{2}x = 1 - \cos^{2}x$

Substitute $\sin^{2}x$ in the equation $4\cos x=-\sin^{2}x + 1$. We get $4\cos x=-(1 - \cos^{2}x)+1$.

Step2: Simplify the equation

Expand the right - hand side: $4\cos x=-1+\cos^{2}x + 1$. Then the equation becomes $\cos^{2}x-4\cos x = 0$.

Step3: Factor the equation

Factor out $\cos x$: $\cos x(\cos x - 4)=0$.

Step4: Solve for $\cos x$

Set each factor equal to zero. For $\cos x=0$, in the interval $[0,2\pi)$, $x=\frac{\pi}{2},\frac{3\pi}{2}$. For $\cos x - 4=0$, i.e., $\cos x = 4$. But since $-1\leqslant\cos x\leqslant1$, $\cos x = 4$ has no solutions.

Answer:

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