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

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

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

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

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

Solution steps

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

Factor

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

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

Solving each part separately

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

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

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

Combine all the solutions

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

sin (2 x) = \frac{( 6 m - 5 )}{5}

4 cos^{2} (x) + 6 = 7

sec (θ) - 1 = (2 - 1) tan (θ)

cos (a) = - \frac{4}{5}

2 tan (θ) - sec (θ) = 0

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