What is the general solution for tan(2x)sin(x)-sin(x)-tan(2x)+1=0 ?

The general solution for tan(2x)sin(x)-sin(x)-tan(2x)+1=0 is x= pi/8+(pin)/2 ,x= pi/2+2pin

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tan(2x)sin(x)-sin(x)-tan(2x)+1=0

tan (2 x) sin (x) - sin (x) - tan (2 x) + 1 = 0

x = \frac{π}{8} + \frac{πn}{2}, x = \frac{π}{2} + 2 πn

Solution steps

tan (2 x) sin (x) - sin (x) - tan (2 x) + 1 = 0

Factor

tan (2 x) sin (x) - sin (x) - tan (2 x) + 1 : (tan (2 x) - 1) (sin (x) - 1)

(tan (2 x) - 1) (sin (x) - 1) = 0

Solving each part separately

tan (2 x) - 1 = 0 or sin (x) - 1 = 0

tan (2 x) - 1 = 0 : x = \frac{π}{8} + \frac{πn}{2}

sin (x) - 1 = 0 : x = \frac{π}{2} + 2 πn

Combine all the solutions

x = \frac{π}{8} + \frac{πn}{2}, x = \frac{π}{2} + 2 πn

3 sec (x) - cos (x) - 2 = 0

- 2 sin (t) = - 2

x, r^{2} = sin (2 x)

3 cos (3 t) = 0

2 cos (x) - 1 = 0, 0 \leq x \leq 2 π

What is the general solution for tan(2x)sin(x)-sin(x)-tan(2x)+1=0 ?
The general solution for tan(2x)sin(x)-sin(x)-tan(2x)+1=0 is x= pi/8+(pin)/2 ,x= pi/2+2pin