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Question Number 77027 by mathocean1 last updated on 02/Jan/20

Please help me to solve it in [- π; 0]

\cos 2 x + \cos x + 1 = \sin 3 x + \sin 2 x + \sin x

E xplain details if possible .

Answered by mr W last updated on 02/Jan/20

\cos 2 x + \cos x + 1 = \sin 3 x + \sin 2 x + \sin x

2 \cos^{2} x + \cos x = 3 \sin x - 4 \sin^{3} x + 2 \sin x \cos x + \sin x

(2 \cos x + 1) \cos x = 2 \sin x \cos x (2 \cos x + 1)

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

\Rightarrow \cos x = 0 \Rightarrow x = - \frac{π}{2}

\Rightarrow \cos x = - \frac{1}{2} \Rightarrow x = - \frac{2 π}{3}

\Rightarrow \sin x = \frac{1}{2} \Rightarrow n o s o l u t i o n w i t h i n [- π, 0]

\Rightarrow s o l u t i o n s a r e : x = - \frac{2 π}{3}, - \frac{π}{2}

Commented by mathocean1 last updated on 02/Jan/20

thank you sir