Solve [0,2pi)

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July 23rd 2012, 09:36 PM
Calcgirl

Solve [0,2pi)

2 similar problems that are really challenging for me.

1) solve [0,2pi) sin 2x-sinx = 0

2) solve [0,2pi) tan^2x sinx= tan^2x

Also how do i rewrite cos^3 x so that it does not contain power greater than 1? I don't think i can use double angle identity since it has power of 3.. not multiples of 2
(Headbang)

Thanks!!
July 25th 2012, 05:13 AM
Amer

Re: Solve [0,2pi)

$\sin 2x - \sin x = 0$
use the identity $\sin 2x = 2\sin x \cos x$ , then take as a factor sinx

$\sin x (2 \cos x - 1 ) = 0$

$\sin x = 0 \Rightarrow x = 0,\pi,2\pi$

$2\cos x - 1 = 0 \Rightarrow \cos x = \frac{1}{2}\;\Rightarrow x = \frac{\pi}{3},\frac{5\pi}{3}$

second one move the left side to the right side then take as a factor tan^2 x same strategy above

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