4sin^2 x + sin 2x = 3 find solution set on x∈(0,2π)


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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
$sin^2 x+2sinxcosx+cos^2 x=3−3sin^2 x+cos^2 x (sinx+cosx)^2 =4cos^2 x sinx+cosx=2cosx sinx=cosx sinx=sin((π/2)−x) x=kπ+(−1)^k ((π/2)−x) 2x=kπ+(π/2) 4x=2kπ+π x=(2k+1)(π/4) {k∈Z so solution x∈[0,2π] are (π/4),((3π)/4) ,((5π)/4),((7π)/4) but at x=((3π)/4) ,((7π)/4) are not valid It has another generic solution x=kπ−((3π)/4) net set∈{(π/4),(π/2)+tan^(−1) (1/3),((5π)/4),((3π)/2)+tan^(−1) (1/3)}$
${s i n}^{2} x + 2 s i n x c o s x + {c o s}^{2} x = 3 - 3 {s i n}^{2} x + {c o s}^{2} x$ ${(s i n x + c o s x)}^{2} = 4 {c o s}^{2} x$ $s i n x + c o s x = 2 c o s x$ $s i n x = c o s x$ $s i n x = s i n (\frac{π}{2} - x)$ $x = k π + {(- 1)}^{k} (\frac{π}{2} - x)$ $2 x = k π + \frac{π}{2}$ $4 x = 2 k π + π$ $x = (2 k + 1) \frac{π}{4} {k \in Z$ $s o s o l u t i o n x \in [0, 2 π] a r e \frac{π}{4}, \frac{3 π}{4}, \frac{5 π}{4}, \frac{7 π}{4} b u t a t x = \frac{3 π}{4}, \frac{7 π}{4} a r e n o t v a l i d$ $I t h a s a n o t h e r g e n e r i c s o l u t i o n x = k π - \frac{3 π}{4}$ $n e t s e t \in {\frac{π}{4}, \frac{π}{2} + \tan^{- 1} \frac{1}{3}, \frac{5 π}{4}, \frac{3 π}{2} + \tan^{- 1} \frac{1}{3}}$
	Answered by MJS last updated on 29/Jun/20
	$4sin^2 x +sin 2x =3 t=tan x ((t^2 +2t−3)/(t^2 +1))=0 t_1 =−3 ⇒ x_1 =arctan (1/3) +(n−(1/2))π t_2 =1 ⇒ x_2 =(π/4)+nπ 0≤x<2π ⇒ x∈{(π/4), (π/2)+arctan (1/3), ((5π)/4), ((3π)/2)+arctan (1/3)}$
	${4 \sin}^{2} x + \sin 2 x = 3$ $t = \tan x$ $\frac{t^{2} + 2 t - 3}{t^{2} + 1} = 0$ $t_{1} = - 3 \Rightarrow x_{1} = \arctan \frac{1}{3} + (n - \frac{1}{2}) π$ $t_{2} = 1 \Rightarrow x_{2} = \frac{π}{4} + n π$ $0 ⩽ x < 2 π \Rightarrow x \in {\frac{π}{4}, \frac{π}{2} + \arctan \frac{1}{3}, \frac{5 π}{4}, \frac{3 π}{2} + \arctan \frac{1}{3}}$
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