Solve for real numbers: 4tan(x) + ((sin(5x))/(cos^5 (x))) = 0


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Question Number 157964 by HongKing last updated on 30/Oct/21

$Solve for real numbers :$ $4 \tan (x) + \frac{\sin (5 x)}{\cos^{5} (x)} = 0$
Commented by tounghoungko last updated on 30/Oct/21
$sin (5x)=5sin x cos^4 x−10sin^3 x cos^2 x+sin^5 x ⇔ ((4sin x cos^4 x+5sin xcos^4 x−10sin^3 xcos^2 x+sin^5 x)/(cos^5 x))=0 ⇔9sin x cos^4 x−10sin^3 x cos^2 x+sin^5 x = 0 sin x(9cos^4 x−10sin^2 x cos^2 x+sin^4 x )=0 (1)sin x =0 ⇒x = { ((2nπ)),(((2n+1)π)) :} ; n∈Z (2)9(((1+cos 2x)/2))^2 −(5/2)sin^2 2x +(((1−cos 2x)/2))^2 = 0 let cos 2x = t ⇒9(((1+t)/2))^2 −(5/2)(1−t^2 )+(((1−t)/2))^2 = 0 ⇒((9t^2 +18t+9)/4)−(((5−5t^2 ))/2) +((1−2t+t^2 )/4) =0 ⇒9t^2 +18t+9−10+10t^2 +1−2t+t^2 =0 ⇒20t^2 +16t = 0 ⇒4t(5t+4)=0 (3) 4t=0⇒4cos 2x = 0 ⇒cos 2x = cos (π/2) ; x= ± (π/4) +nπ (4) t=−(4/5)⇒cos 2x =−(4/5) ⇒2x = ± arccos (−(4/5))+2nπ ⇒x=± (1/2) arccos (−(4/5))+nπ$
$\sin (5 x) = 5 \sin x \cos^{4} x - 10 \sin^{3} x \cos^{2} x + \sin^{5} x$ $\Leftrightarrow \frac{4 \sin x \cos^{4} x + 5 \sin x \cos^{4} x - 10 \sin^{3} x \cos^{2} x + \sin^{5} x}{\cos^{5} x} = 0$ $\Leftrightarrow 9 \sin x \cos^{4} x - 10 \sin^{3} x \cos^{2} x + \sin^{5} x = 0$ $\sin x (9 \cos^{4} x - 10 \sin^{2} x \cos^{2} x + \sin^{4} x) = 0$ $(1) \sin x = 0 \Rightarrow x = {\begin{cases} 2 n π \\ (2 n + 1) π \end{cases}; n \in Z$ $(2) 9 {(\frac{1 + \cos 2 x}{2})}^{2} - \frac{5}{2} \sin^{2} 2 x + {(\frac{1 - \cos 2 x}{2})}^{2} = 0$ $l e t \cos 2 x = t$ $\Rightarrow 9 {(\frac{1 + t}{2})}^{2} - \frac{5}{2} (1 - t^{2}) + {(\frac{1 - t}{2})}^{2} = 0$ $\Rightarrow \frac{9 t^{2} + 18 t + 9}{4} - \frac{(5 - 5 t^{2})}{2} + \frac{1 - 2 t + t^{2}}{4} = 0$ $\Rightarrow 9 t^{2} + 18 t + 9 - 10 + 10 t^{2} + 1 - 2 t + t^{2} = 0$ $\Rightarrow 20 t^{2} + 16 t = 0$ $\Rightarrow 4 t (5 t + 4) = 0$ $(3) 4 t = 0 \Rightarrow 4 \cos 2 x = 0$ $\Rightarrow \cos 2 x = \cos \frac{π}{2}; x = \pm \frac{π}{4} + n π$ $(4) t = - \frac{4}{5} \Rightarrow \cos 2 x = - \frac{4}{5}$ $\Rightarrow 2 x = \pm \arccos (- \frac{4}{5}) + 2 n π$ $\Rightarrow x = \pm \frac{1}{2} \arccos (- \frac{4}{5}) + n π$
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