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

The general solution for 1+sin(x)=cos(2x) is x=2pin,x=pi+2pin,x=(7pi)/6+2pin,x=(11pi)/6+2pin

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

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

x = 2 πn, x = π + 2 πn, x = \frac{7 π}{6} + 2 πn, x = \frac{11 π}{6} + 2 πn

Solution steps

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

Subtract

cos (2 x)

from both sides

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

Rewrite using trig identities

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

Solve by substitution

sin (x) = 0, sin (x) = - \frac{1}{2}

sin (x) = 0 : x = 2 πn, x = π + 2 πn

sin (x) = - \frac{1}{2} : x = \frac{7 π}{6} + 2 πn, x = \frac{11 π}{6} + 2 πn

Combine all the solutions

x = 2 πn, x = π + 2 πn, x = \frac{7 π}{6} + 2 πn, x = \frac{11 π}{6} + 2 πn

tan (4 5^{\circ} + x) + tan (4 5^{\circ} - x) = 4

1 - cos (θ) = sin (θ)

3 cot (\frac{3 π}{7} x) + 3 = 0

sin (u) = 1

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

What is the general solution for 1+sin(x)=cos(2x) ?
The general solution for 1+sin(x)=cos(2x) is x=2pin,x=pi+2pin,x=(7pi)/6+2pin,x=(11pi)/6+2pin