3 cos x = 13 sin (((2x)/3)) + 17 cos ((x/3))


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Question Number 136121 by bramlexs22 last updated on 18/Mar/21

$3 \cos x = 13 \sin (\frac{2 x}{3}) + 17 \cos (\frac{x}{3})$
	Answered by EDWIN88 last updated on 19/Mar/21
	$let (x/3) = t ⇒3cos 3t = 13sin 2t+17cos t ⇒3(4cos^3 t−3cos t)=26sin t cos t +17cos t 12cos^3 t−26cos t−26sin t cos t = 0 2cos t (6cos^2 t−13sin t−13)=0 (∗) 2cos t = 0 ⇒t=±(π/2)+2nπ (x/3) = ± (π/2)+2nπ ⇒x = ± ((3π)/2)+6nπ (∗)⇒6−6sin^2 t−13sin t−13 =0 ⇒ 6sin^2 t+13sin t+7=0 ⇒(6sin t+7)(sin t+1)=0 { ((sin t =−(7/6) ⇒t∉R)),((sin t=−1⇒t=−(π/2)+2nπ , (x/3)=−(π/2)+2nπ)) :}$
	$let \frac{x}{3} = t \Rightarrow 3 \cos 3 t = 13 \sin 2 t + 17 \cos t$ $\Rightarrow 3 (4 \cos^{3} t - 3 \cos t) = 26 \sin t \cos t + 17 \cos t$ $12 \cos^{3} t - 26 \cos t - 26 \sin t \cos t = 0$ $2 \cos t (6 \cos^{2} t - 13 \sin t - 13) = 0$ $() 2 \cos t = 0 \Rightarrow t = \pm \frac{π}{2} + 2 n π$ $\frac{x}{3} = \pm \frac{π}{2} + 2 n π \Rightarrow x = \pm \frac{3 π}{2} + 6 n π$ $() \Rightarrow 6 - 6 \sin^{2} t - 13 \sin t - 13 = 0$ $\Rightarrow 6 \sin^{2} t + 13 \sin t + 7 = 0$ $\Rightarrow (6 \sin t + 7) (\sin t + 1) = 0$ ${\begin{cases} \sin t = - \frac{7}{6} \Rightarrow t \notin R \\ \sin t = - 1 \Rightarrow t = - \frac{π}{2} + 2 n π, \frac{x}{3} = - \frac{π}{2} + 2 n π \end{cases}$
	Commented by EDWIN88 last updated on 19/Mar/21

	$yes . thanks you$
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