What is the general solution for 1+sin(x)=2cos^2(x) ?

The general solution for 1+sin(x)=2cos^2(x) is x= pi/6+2pin,x=(5pi)/6+2pin,x=(3pi)/2+2pin

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1+sin(x)=2cos^2(x)

1 + sin (x) = 2 cos^{2} (x)

x = \frac{π}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn, x = \frac{3 π}{2} + 2 πn

Solution steps

1 + sin (x) = 2 cos^{2} (x)

Subtract

2 cos^{2} (x)

from both sides

1 + sin (x) - 2 cos^{2} (x) = 0

Rewrite using trig identities

- 1 + sin (x) + 2 sin^{2} (x) = 0

Solve by substitution

sin (x) = \frac{1}{2}, sin (x) = - 1

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

sin (x) = - 1 : x = \frac{3 π}{2} + 2 πn

Combine all the solutions

x = \frac{π}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn, x = \frac{3 π}{2} + 2 πn

3 cot (x) = 1

tan (x) = \frac{5}{12}

cos (x) = - \frac{3}{5}

tan (x) = - \frac{1}{3}

6 sin^{2} (θ) - sin (θ) - 2 = 0

What is the general solution for 1+sin(x)=2cos^2(x) ?
The general solution for 1+sin(x)=2cos^2(x) is x= pi/6+2pin,x=(5pi)/6+2pin,x=(3pi)/2+2pin