Question #20cb6

1 Answer
Jan 26, 2018

# x in {0, pi/3, 2pi/3, pi, 4pi/3, 5pi/3, 2pi}#.

Explanation:

#4cos^2x-sin^2 2x+color(red)(5)sin^2x=4#.

#:. 4cos^2x-sin^2 2x+color(red)((4+1))sin^2x=4#.

#:. 4cos^2x-sin^2 2x+4sin^2x+sin^2x=4#.

#:. cancel((4cos^2x+4sin^2x))-sin^2 2x+sin^2x=cancel4#.

#:. sin^2x-sin^2 2x=0#.

But, #sin^2theta-sin^2phi=sin(theta+phi)sin(theta-phi)#.

#:. sin(x+2x)sin(x-2x)=0, i.e., #

# (sin3x)(-sinx)=0..............[because, sin(-x)=-sinx]#,

#:. sin3x=0, or, sinx=0#.

#:. 3x=kpi, or x=kpi, k in ZZ#.

#:. x=kpi/3, or x=kpi, k in ZZ#.

#:. x in {kpi/3}uu{kpi}, k in ZZ#.

But, #because {kpi} sub {kpi/3}, k in ZZ, x in {kpi/3 | k in ZZ}#.

#:. x in [0,2pi] sup {kpi/3 | k in ZZ}#

#={0, pi/3, 2pi/3, pi, 4pi/3, 5pi/3, 2pi}#.