Honestly, have no idea, I know that the $\cos2x$ turns into $1-\sin x$, but after that i'm not sure. Help is greatly appreciated
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2Hint: $\sin^2+\cos^2=1$. – Jonatan B. Bastos Sep 22 '17 at 03:50
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$4 \cos^2 x − \sin^2 (2x) + 5 \sin^2 x = 4$; $ \sin^2 x − \sin^2 (2x) = 0$, $$ (\sin x − \sin 2x)(\sin x + \sin 2x) = 0$$ $$\sin^2x(1-2 \cos x)(1+2 \cos x)=0$$ Can you finish?
Vasili
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Use https://math.stackexchange.com/questions/175143/prove-sinab-sina-b-sin2a-sin2b – lab bhattacharjee Sep 22 '17 at 04:05
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$$4-4\sin^2x\cos^2x+\sin^2x=4,$$ which gives $\sin{x}=0$ and $\{0,\pi,2\pi\}$
or $$4\cos^2x-1=0$$ or $$2(1+\cos2x)-1=0$$ or $$\cos2x=-\frac{1}{2},$$ which gives $$\left\{\frac{2\pi}{3},\frac{\pi}{3},\frac{4\pi}{3},\frac{5\pi}{3}\right\}$$
Michael Rozenberg
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Compute this,
$$4(1-\sin^2(x)) - 4\sin^2(x)(1-\sin^2(x))+5\sin^2(x) = 4, x \in [0, 2\pi]$$
Then, you get.
$$\sin^2(x)(4\sin^2(x)-3) = 0$$
$$\sin^2(x) = \frac34, 0$$
$$\sin(x) = \pm\frac{\sqrt 3}2, 0$$
which gives, $$ x = [0, \frac\pi3, \frac{2\pi}3, \pi, \frac{4\pi}3, \frac{5\pi}3, 2\pi]$$
kimi Tanaka
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Type \sin x, \cos x, \tan x, \csc x, \sec x, and \cot x, respectively, when you are in math mode to produce $\sin x$, $\cos x$, $\tan x$, $\csc x$, $\sec x$, and $\cot x$. – N. F. Taussig Sep 22 '17 at 09:54
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$4 \cos^2 x − \sin^2 (2x) + 5 \sin^2 x = 4$
$4 \cos^2 x + 4 \sin^2 x− \sin^2 (2x) +\sin^2 x = 4$
$4 − \sin^2 (2x) +\sin^2 x = 4$
$− \sin^2 (2x) +\sin^2 x = 0$
$-2 \sin^2 x\cos^2 x +\sin^2 x=0 $
$\sin^2 x \left(\cos^2 x -\frac{1}{2} \right)=0.$
$x=0\vee x=\pi\vee x=2\pi \vee x=\frac{\pi}{4}\vee x=\frac{3\pi}{4}\vee x=\frac{5\pi}{4}\vee x=\frac{7\pi}{4}$
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1it should be 4 not 2 with $sin(2a)$ in $-2 \sin^2 x\cos^2 x +\sin^2 x=0$ – user577215664 Sep 22 '17 at 04:09