sin-2-4x-cos-2-x-2sin-4x-cos-2-x-

Question Number 135867 by liberty last updated on 16/Mar/21

\sin^{2} (4 x) + \cos^{2} (x) = 2 \sin (4 x) \cos^{2} (x)

Answered by EDWIN88 last updated on 16/Mar/21

recall \cos^{2} x = \frac{1}{2} + \frac{1}{2} \cos 2 x

\Leftrightarrow {[2 \sin 2 x \cos 2 x]}^{2} + \frac{1}{2} + \frac{1}{2} \cos 2 x = 4 \sin 2 x \cos 2 x (\frac{1}{2} + \frac{1}{2} \cos 2 x)

4 \sin^{2} 2 x \cos^{2} 2 x + \frac{1}{2} \cos 2 x + \frac{1}{2} = 2 \sin 2 x \cos 2 x + 2 \sin 2 x \cos^{2} 2 x

8 \sin^{2} 2 x \cos^{2} 2 x + \cos 2 x + 1 = 4 \sin 2 x \cos 2 x + 4 \sin 2 x \cos^{2} 2 x

8 \sin^{2} 2 x \cos^{2} 2 x + \cos 2 x + \sin^{2} 2 x + \cos^{2} 2 x = 4 \sin 2 x \cos 2 x + 4 \sin 2 x \cos^{2} 2 x

Answered by liberty last updated on 17/Mar/21

Answered by MJS_new last updated on 17/Mar/21

using trigonometric formulas I get

[\sin x = s \land \cos x = c]

- c^{2} (64 c^{2} s^{4} - 16 s^{2} - 1) = 8 c^{3} s (2 c^{2} - 1)

\Rightarrow \cos x = 0 \Leftrightarrow ★ x = n π - \frac{π}{2} \land n \in Z ★

- (64 c^{2} s^{4} - 16 s^{2} - 1) = 8 c s (2 c^{2} - 1)

this has no real solution

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