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)}