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=(3pi)/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{3 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{3 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{3 π}{4} + πn, x = \frac{π}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn

sin (θ) = \frac{0.3924}{cos ( θ )}

x, y = 3 sin (x)

6 - 4 cos^{2} (x) - 5 cos (x) = 0

cos (x) = \frac{32}{34.54}

tan (a) = \frac{51}{140}

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=(3pi)/4+pin,x= pi/6+2pin,x=(5pi)/6+2pin