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

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

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Answer 1 · 2016-12-22T14:53:38

${0 + 2pin, pi/2 + 2pin, pi + 2pin, (3pi)/2 + 2pin}$

Explanation:

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

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

$sin2xcos2x + sin2xcos2x + 2sin2x = 0$

$2sin2xcos2x + 2sin2x = 0$

$2sin2x(cos2x + 1) = 0$

Case 1: $2sin2x= 0$

Use $sin2theta = 2sinthetacostheta$ :

$2(2sinxcosx) = 0$

$4sinxcosx= 0$

Whenever $sinx = 0$ and whenever $cosx = 0$ , the entire equation will equal $0$ . So, $x = 0 + 2pin$ , $pi/2 + 2pin$ , $pi + 2pin$ , (3pi)/2 + 2pin#.

Case 2: $cos2x + 1 = 0$

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

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

$2cos^2x = 0$

$cos^2x= 0$

$cosx =0$

$x = pi/2 + 2pin$ and $(3pi)/2 + 2pin$

Hopefully this helps!

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