Solve for real numbers: tan(10x) = tan^5 (2x)


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Question Number 157489 by MathSh last updated on 23/Oct/21

$Solve for real numbers :$ $\tan (10 x) = \tan^{5} (2 x)$
	Answered by MJS_new last updated on 24/Oct/21

	$let t = \tan 2 x$ $\frac{t (t^{4} - 10 t^{2} + 5)}{5 t^{4} - 10 t^{2} + 1} = t^{5}$ ${(t + 1)}^{3} t {(t - 1)}^{3} (t^{2} + 1) = 0$ $t = - 1 \lor t = 0 \lor t = 1$ $x = - \frac{π}{8} + \frac{n π}{2} \lor x = \frac{n π}{2} \lor x = \frac{π}{8} + \frac{n π}{2}$
	Commented by tounghoungko last updated on 24/Oct/21

	$\tan 5 x = \frac{\tan^{5} x - 10 \tan^{3} x + 5 \tan x}{1 - 10 \tan^{2} x + 5 \tan^{4} x}$
	Commented by MJS_new last updated on 24/Oct/21

	$yes$
	Commented by MathSh last updated on 24/Oct/21

	$Thank you dear S er very nice$
	Answered by tounghoungko last updated on 24/Oct/21
	$Solve for x∈R . tan (10x)=tan^5 (2x) ⇔ tan (5θ)=((sin^5 θ)/(cos^5 θ)) ,[θ=2x] ⇔((sin 5θ)/(cos 5θ)) = ((sin^5 θ)/(cos^5 θ)) ⇔ ((5sin θ−20sin^3 θ+16sin^5 θ)/(5cos θ−20cos^3 θ+16cos^5 θ))=((sin^5 θ)/(cos^5 θ)) ⇔5sin θcos^5 θ−20sin^3 θcos^5 θ+16sin^5 θcos^5 θ= 5cos θsin^5 θ−20cos^3 θsin^5 θ+16cos^5 θsin^5 θ ⇔sin θcos θ [5cos^4 θ−20sin^2 θcos^4 θ+16sin^4 θcos^4 θ− [5sin^4 θ+20cos^2 θsin^4 θ−16sin^4 θcos^4 θ ]=0 ⇔ (1/2)sin 2θ [5(cos^4 θ−sin^4 θ)+20(cos^2 θsin^4 θ−sin^2 θcos^4 θ)]=0 ⇔(1/2)sin 2θ [5(cos^2 θ−sin^2 θ)+20cos^2 θsin^2 θ(sin^2 θ−cos^2 θ)]=0 ⇔(5/2)sin 2θ cos 2θ (1−sin^2 2θ)=0 ⇔(5/4)sin 4θ (1−sin^2 2θ)=0 (i) (5/4)sin 4θ=0⇒ { ((4θ=2nπ)),((4θ=(1+2n)π)) :} ⇒ { ((8x=2nπ)),((8x=(1+2n)π)) :}⇒ { ((x=((nπ)/4))),((x=(((1+2n)/8))π)) :} (ii) 1−sin^2 2θ=0 ⇒ { ((sin 2θ=1)),((sin 2θ=−1)) :} ⇒ { (( 2θ=(π/2)+2kπ)),((2θ=((3π)/2)+2kπ)) :} { ((4x=(π/2)+2kπ)),((4x=((3π)/2)+2kπ)) :}⇒ { ((x=(((4k+1)/8))π)),((x=(((3+4k)/8))π)) :}$
	$S o l v e f o r x \in R . \tan (10 x) = \tan^{5} (2 x)$ $\Leftrightarrow \tan (5 θ) = \frac{\sin^{5} θ}{\cos^{5} θ}, [θ = 2 x]$ $\Leftrightarrow \frac{\sin 5 θ}{\cos 5 θ} = \frac{\sin^{5} θ}{\cos^{5} θ}$ $\Leftrightarrow \frac{5 \sin θ - 20 \sin^{3} θ + 16 \sin^{5} θ}{5 \cos θ - 20 \cos^{3} θ + 16 \cos^{5} θ} = \frac{\sin^{5} θ}{\cos^{5} θ}$ $\Leftrightarrow 5 \sin θ \cos^{5} θ - 20 \sin^{3} θ \cos^{5} θ + 16 \sin^{5} θ \cos^{5} θ =$ $5 \cos θ \sin^{5} θ - 20 \cos^{3} θ \sin^{5} θ + 16 \cos^{5} θ \sin^{5} θ$ $\Leftrightarrow \sin θ \cos θ [5 \cos^{4} θ - 20 \sin^{2} θ \cos^{4} θ + 16 \sin^{4} θ \cos^{4} θ -$ $[5 \sin^{4} θ + 20 \cos^{2} θ \sin^{4} θ - 16 \sin^{4} θ \cos^{4} θ] = 0$ $\Leftrightarrow \frac{1}{2} \sin 2 θ [5 (\cos^{4} θ - \sin^{4} θ) + 20 (\cos^{2} θ \sin^{4} θ - \sin^{2} θ \cos^{4} θ)] = 0$ $\Leftrightarrow \frac{1}{2} \sin 2 θ [5 (\cos^{2} θ - \sin^{2} θ) + 20 \cos^{2} θ \sin^{2} θ (\sin^{2} θ - \cos^{2} θ)] = 0$ $\Leftrightarrow \frac{5}{2} \sin 2 θ \cos 2 θ (1 - \sin^{2} 2 θ) = 0$ $\Leftrightarrow \frac{5}{4} \sin 4 θ (1 - \sin^{2} 2 θ) = 0$ $(i) \frac{5}{4} \sin 4 θ = 0 \Rightarrow {\begin{cases} 4 θ = 2 n π \\ 4 θ = (1 + 2 n) π \end{cases}$ $\Rightarrow {\begin{cases} 8 x = 2 n π \\ 8 x = (1 + 2 n) π \end{cases} \Rightarrow {\begin{cases} x = \frac{n π}{4} \\ x = (\frac{1 + 2 n}{8}) π \end{cases}$ $(i i) 1 - \sin^{2} 2 θ = 0$ $\Rightarrow {\begin{cases} \sin 2 θ = 1 \\ \sin 2 θ = - 1 \end{cases} \Rightarrow {\begin{cases} 2 θ = \frac{π}{2} + 2 k π \\ 2 θ = \frac{3 π}{2} + 2 k π \end{cases}$ ${\begin{cases} 4 x = \frac{π}{2} + 2 k π \\ 4 x = \frac{3 π}{2} + 2 k π \end{cases} \Rightarrow {\begin{cases} x = (\frac{4 k + 1}{8}) π \\ x = (\frac{3 + 4 k}{8}) π \end{cases}$
	Commented by MathSh last updated on 24/Oct/21

	$Thank you dear S er cool$
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