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

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

Commented byDwaipayan 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)}

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)}