How do you solve $sin 4 x + 2 sin 2 x = 0$ $sin4x+2sin2x=0$ ?

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How do you solve $sin 4 x + 2 sin 2 x = 0$ $sin4x+2sin2x=0$ ?

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Answer 1 · 2016-12-22T14:53:38

${0 + 2 π n, \frac{π}{2} + 2 π n, π + 2 π n, \frac{3 π}{2} + 2 π n}$ ${0 + 2pin, pi/2 + 2pin, pi + 2pin, (3pi)/2 + 2pin}$

Explanation:

Expand using $sin (A + B) = sin A cos B + cos A sin B$ $sin(A + B) = sinAcosB + cosAsinB$ :

$sin (2 x + 2 x) + 2 sin 2 x = 0$ $sin(2x + 2x) + 2sin2x= 0$

$sin 2 x cos 2 x + sin 2 x cos 2 x + 2 sin 2 x = 0$ $sin2xcos2x + sin2xcos2x + 2sin2x = 0$

$2 sin 2 x cos 2 x + 2 sin 2 x = 0$ $2sin2xcos2x + 2sin2x = 0$

$2 sin 2 x (cos 2 x + 1) = 0$ $2sin2x(cos2x + 1) = 0$

Case 1: $2 sin 2 x = 0$ $2sin2x= 0$

Use $sin 2 θ = 2 sin θ cos θ$ $sin2theta = 2sinthetacostheta$ :

$2 (2 sin x cos x) = 0$ $2(2sinxcosx) = 0$

$4 sin x cos x = 0$ $4sinxcosx= 0$

Whenever $sin x = 0$ $sinx = 0$ and whenever $cos x = 0$ $cosx = 0$ , the entire equation will equal $0$ $0$ . So, $x = 0 + 2 π n$ $x = 0 + 2pin$ , $\frac{π}{2} + 2 π n$ $pi/2 + 2pin$ , $π + 2 π n$ $pi + 2pin$ , (3pi)/2 + 2pin#.

Case 2: $cos 2 x + 1 = 0$ $cos2x + 1 = 0$

Use the identity $cos 2 x = 2 {cos}^{2} x - 1$ $cos2x = 2cos^2x - 1$ .

$2 {cos}^{2} x - 1 + 1 = 0$ $2cos^2x - 1 + 1 = 0$

$2 {cos}^{2} x = 0$ $2cos^2x = 0$

${cos}^{2} x = 0$ $cos^2x= 0$

$cos x = 0$ $cosx =0$

$x = \frac{π}{2} + 2 π n$ $x = pi/2 + 2pin$ and $\frac{3 π}{2} + 2 π n$ $(3pi)/2 + 2pin$

Hopefully this helps!

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How do you solve $sin 4 x + 2 sin 2 x = 0$ $sin4x+2sin2x=0$ ?

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