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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?\n
Commented bymr W last updated on 17/Mar/21

$2 \cos^{3} x + \sin x - 3 \sin^{2} x \cos x = 0$ $2 \cos^{3} x + \sin x - 3 (1 - \cos^{2} x) \cos x = 0$ $5 \cos^{3} x + \sin x - 3 \cos x = 0$ $\cos x (5 \cos^{2} x - 3) + \sin x = 0$ $3 - \frac{5}{\tan^{2} x + 1} = \tan x$ $3 - \frac{5}{t^{2} + 1} = t$ $t^{3} - 3 t^{2} + t + 2 = 0$ $(t - 2) (t^{2} - t - 1) = 0$ $\Rightarrow t = \tan x = 2 \Rightarrow x = n π + {t a n}^{- 1} 2$ $\Rightarrow t = \tan x = \frac{1 \pm \sqrt{5}}{2} \Rightarrow x = n π + {t a n}^{- 1} \frac{1 \pm \sqrt{5}}{2}$
Commented byliberty last updated on 17/Mar/21
$replacing sin x=sin ^3 x+sin x cos ^2 x (•) 2cos ^3 x+sin ^3 x+sin xcos ^2 x−3sin ^2 x cos x = 0 divided by cos ^3 x 2+tan ^3 x+tan x−3tan ^2 x = 0 ⇒tan ^3 x−3tan ^2 x+tan x+2 = 0 let tan x = u ⇒u^3 −3u^2 +u+2 = 0 (u−2)(u^2 −u−1)=0 → { ((u=2⇒tan x=2)),((u = ((1±(√5))/2)⇒tan x=((1±(√5))/2))) :}$
$r e p l a c i n g \sin x = \sin^{3} x + \sin x \cos^{2} x$ $(∙) 2 \cos^{3} x + \sin^{3} x + \sin x \cos^{2} x - 3 \sin^{2} x \cos x = 0$ $d i v i d e d b y \cos^{3} x$ $2 + t a n^{3} x + t a n x - 3 t a n^{2} x = 0$ $\Rightarrow t a n^{3} x - 3 t a n^{2} x + t a n x + 2 = 0$ $l e t t a n x = u$ $\Rightarrow u^{3} - 3 u^{2} + u + 2 = 0$ $Extra close brace or missing open brace$ $\to {\begin{cases} u = 2 \Rightarrow t a n x = 2 \\ u = \frac{1 \pm \sqrt{5}}{2} \Rightarrow t a n x = \frac{1 \pm \sqrt{5}}{2} \end{cases}$
	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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