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

The general solution for (tan(x)-1)(2sin(x)-1)=0 is x= pi/4+pin,x= pi/6+2pin,x=(5pi)/6+2pin

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

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

x = \frac{π}{4} + πn, x = \frac{π}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn

Solution steps

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

Solving each part separately

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

tan (x) - 1 = 0 : x = \frac{π}{4} + πn

2 sin (x) - 1 = 0 : x = \frac{π}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn

Combine all the solutions

x = \frac{π}{4} + πn, x = \frac{π}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn

cos (x) = \frac{20}{29}

2 cot^{4} (x) - cot^{2} (x) - 15 = 0

cos (θ) = \frac{11}{6}

1 + cos (x) = 3 cos (x)

2 cot (θ) + 1 = 0

What is the general solution for (tan(x)-1)(2sin(x)-1)=0 ?
The general solution for (tan(x)-1)(2sin(x)-1)=0 is x= pi/4+pin,x= pi/6+2pin,x=(5pi)/6+2pin