4sin-2-x-sin-2x-3-find-solution-set-on-x-0-2pi-

Question Number 100988 by bobhans last updated on 29/Jun/20

4 \sin^{2} x + \sin 2 x = 3

f i n d s o l u t i o n s e t o n x \in (0, 2 π)

Commented by Dwaipayan Shikari last updated on 29/Jun/20

{s i n}^{2} x + 2 s i n x c o s x + {c o s}^{2} x = 3 - 3 {s i n}^{2} x + {c o s}^{2} x

{(s i n x + c o s x)}^{2} = 4 {c o s}^{2} x

s i n x + c o s x = 2 c o s x

s i n x = c o s x

s i n x = s i n (\frac{π}{2} - x)

x = k π + {(- 1)}^{k} (\frac{π}{2} - x)

2 x = k π + \frac{π}{2}

4 x = 2 k π + π

x = (2 k + 1) \frac{π}{4} {k \in Z

s o s o l u t i o n x \in [0, 2 π] a r e \frac{π}{4}, \frac{3 π}{4}, \frac{5 π}{4}, \frac{7 π}{4} b u t a t x = \frac{3 π}{4}, \frac{7 π}{4} a r e n o t v a l i d

I t h a s a n o t h e r g e n e r i c s o l u t i o n x = k π - \frac{3 π}{4}

n e t s e t \in {\frac{π}{4}, \frac{π}{2} + \tan^{- 1} \frac{1}{3}, \frac{5 π}{4}, \frac{3 π}{2} + \tan^{- 1} \frac{1}{3}}

Answered by MJS last updated on 29/Jun/20

{4 \sin}^{2} x + \sin 2 x = 3

t = \tan x

\frac{t^{2} + 2 t - 3}{t^{2} + 1} = 0

t_{1} = - 3 \Rightarrow x_{1} = \arctan \frac{1}{3} + (n - \frac{1}{2}) π

t_{2} = 1 \Rightarrow x_{2} = \frac{π}{4} + n π

0 ⩽ x < 2 π \Rightarrow x \in {\frac{π}{4}, \frac{π}{2} + \arctan \frac{1}{3}, \frac{5 π}{4}, \frac{3 π}{2} + \arctan \frac{1}{3}}

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