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Question Number 135974 by liberty last updated on 17/Mar/21

How do you solve for 2cos^3(x) + sinx – 3sin^2 (x) cos(x) =0?

Commented by mr W last updated on 17/Mar/21

2 \cos^{3} x + \sin x - 3 \sin^{2} x \cos x = 0

2 \cos^{3} x + \sin x - 3 (1 - \cos^{2} x) \cos x = 0

5 \cos^{3} x + \sin x - 3 \cos x = 0

\cos x (5 \cos^{2} x - 3) + \sin x = 0

3 - \frac{5}{\tan^{2} x + 1} = \tan x

3 - \frac{5}{t^{2} + 1} = t

t^{3} - 3 t^{2} + t + 2 = 0

(t - 2) (t^{2} - t - 1) = 0

\Rightarrow t = \tan x = 2 \Rightarrow x = n π + \tan^{- 1} 2

\Rightarrow t = \tan x = \frac{1 \pm \sqrt{5}}{2} \Rightarrow x = n π + \tan^{- 1} \frac{1 \pm \sqrt{5}}{2}

Commented by liberty last updated on 17/Mar/21

r e p l a c i n g \sin x = \sin^{3} x + \sin x \cos^{2} x

(∙) 2 \cos^{3} x + \sin^{3} x + \sin x \cos^{2} x - 3 \sin^{2} x \cos x = 0

d i v i d e d b y \cos^{3} x

2 + \tan^{3} x + \tan x - 3 \tan^{2} x = 0

\Rightarrow \tan^{3} x - 3 \tan^{2} x + \tan x + 2 = 0

l e t \tan x = u

\Rightarrow u^{3} - 3 u^{2} + u + 2 = 0

(u - 2) (u^{2} - u - 1) = 0

\to {\begin{cases} u = 2 \Rightarrow \tan x = 2 \\ u = \frac{1 \pm \sqrt{5}}{2} \Rightarrow \tan x = \frac{1 \pm \sqrt{5}}{2} \end{cases}

Answered by MJS_new last updated on 17/Mar/21

let t = \tan \frac{x}{2}

- \frac{2 (t^{2} + t - 1) (t^{4} - 2 t^{3} - 6 t^{3} + 2 g + 1)}{{(t^{2} + 1)}^{3}} = 0

(t^{2} + t - 1) (t^{2} - (1 + \sqrt{5}) t - 1) (t^{2} - (1 - \sqrt{5}) t - 1) = 0

solve this for t and then x = 2 π n + 2 \arctan t

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