Question

Solve the trigonometric equation given by 

(2sin(x) - 1)(tan(x) - 1) = 0 for 0 ≤ x ≤ 2 pi

Answer 1

LHS= 2sinx - 1 * tanx - 1
=2* sinx - 1 * (sinx/cosx) - 1
=2* sinx - 1* sinx-cosx/cosx
=(2/cosx) - 1 * (sinx-cosx/cosx)
=(2-cosx/cosx) * (sinx-cosx/cosx)
=2sinx-2cosx-cosx*sinx+cos^2x/cos^2x
=2sinx-3cosx*sinx+cos^2x*sin^2x[cos^2x=1/sin^2x)
=2sinx-3cosx*sinx+1
=2sin^2x-3cosx*sinx+1
=2sin^2x-3*(cosx/cosx)+1
=2sin^2x-3+1
=2sin^2x-2
=2(sin^2x-1)
Here if x=90°
Then,
2(sin^2 90°-1)
=2(1-1)
=2(0 <= x <= 2)