(1+sin (π/7))^(3−cos 2x) = (sin (π/(14))+cos (π/(14)))^(10 sin x) find solution


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Question Number 78785 by jagoll last updated on 20/Jan/20

${(1 + \sin \frac{π}{7})}^{3 - \cos 2 x} = {(\sin \frac{π}{14} + \cos \frac{π}{14})}^{10 \sin x}$ $find solution$
	Answered by john santu last updated on 20/Jan/20
	$consider 1+sin (π/7)=(sin (π/(14))+cos (π/(14)))^2 ⇒(sin (π/(14))+cos (π/(14)))^(6−2cos 2x) = (sin (π/(14))+cos (π/(14)))^(10 sin x) ⇒6 − 2cos 2x = 10 sin x −2(1−2sin^2 x)+6−10sin x=0 4sin^2 x−10sin x+4=0 2sin^2 x−5sin x+2=0 (2sin x−1)(sin x−2)=0 sin x=(1/2) { ((x=(π/6)+2nπ)),((x=((5π)/6)+2nπ)) :}$
	$c o n s i d e r 1 + \sin \frac{π}{7} = {(\sin \frac{π}{14} + \cos \frac{π}{14})}^{2}$ $\Rightarrow {(\sin \frac{π}{14} + \cos \frac{π}{14})}^{6 - 2 \cos 2 x} = {(\sin \frac{π}{14} + \cos \frac{π}{14})}^{10 \sin x}$ $\Rightarrow 6 - 2 \cos 2 x = 10 \sin x$ $- 2 (1 - 2 \sin^{2} x) + 6 - 10 \sin x = 0$ $4 \sin^{2} x - 10 \sin x + 4 = 0$ $2 \sin^{2} x - 5 \sin x + 2 = 0$ $(2 \sin x - 1) (\sin x - 2) = 0$ $\sin x = \frac{1}{2} {\begin{cases} x = \frac{π}{6} + 2 n π \\ x = \frac{5 π}{6} + 2 n π \end{cases}$
	Commented by jagoll last updated on 20/Jan/20

	$thanks sir$
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