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Algebra Difficulty 4.9 AIME Find the answer

Find all xx between −π2-\frac{\pi}{2} and π2\frac{\pi}{2} such that 1−sin⁡4x−cos⁡2x=1161-\sin ^{4} x-\cos ^{2} x=\frac{1}{16}.

A number or a short expression. Spacing and $ signs are ignored.

Solution

1−sin⁡4x−cos⁡2x=116⇒(16−16cos⁡2x)−sin⁡4x−1=0⇒16sin⁡4x−1-\sin ^{4} x-\cos ^{2} x=\frac{1}{16} \Rightarrow\left(16-16 \cos ^{2} x\right)-\sin ^{4} x-1=0 \Rightarrow 16 \sin ^{4} x- 16sin⁡2x+1=016 \sin ^{2} x+1=0. Use the quadratic formula in sin⁡x\sin x to obtain sin⁡2x=12±34\sin ^{2} x=\frac{1}{2} \pm \frac{\sqrt{3}}{4}. Since cos⁡2x=1−2sin⁡2x=±32\cos 2 x=1-2 \sin ^{2} x= \pm \frac{\sqrt{3}}{2}, we get x=±π12,±5π12x= \pm \frac{\pi}{12}, \pm \frac{5 \pi}{12}.

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