SOLVE THE FOLLOWING VALUE OF X IN THE INTERVAL [0,2pi)

Question

sinxtanx+sinx=0

Answer 1

Convert tan x to (sin x)/(cos x)

 

(sin x)(sin x)/(cos x) + sin x = 0

 

(sin²x)/(cos x) = - sin x

 

sin²x = (-sin x)(cos x)

 

sin²x + (sin x)(cos x) = 0

 

sin x(sin x + cos x) = 0

 

sin x = 0 and sin x + cos x = 0

 

sin x = 0 when x = 0 and x = π in the interval [0, 2π)

 

sin x + cos x = 0

 

cos x = -sin x

 

1 = (-sin x)/(cos x)

 

-tan x = 1

 

tan x = -1

 

Tangent is negative in quadrants II and IV, and tan x = 1 when x = π/4

 

The reference angles for x = π/4 in quadrants II and IV are 3π/4 and 7π/4, respectively.

 

The solutions to this equation are x = 0, π, 3π/4, and 7π/4