What is the general solution for solvefor x,sin(x)=a-2*q*cos(2x) ?

The general solution for solvefor x,sin(x)=a-2*q*cos(2x) is

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solvefor x,sin(x)=a-2qcos(2x)

solve for

x, sin (x) = a - 2 \cdot q \cdot cos (2 x)

x = arcsin (- \frac{- 1 + 1 + 16 q ( - a + 2 q )}{8 q}) + 2 πn, x = π + arcsin (\frac{- 1 + 1 + 16 q ( - a + 2 q )}{8 q}) + 2 πn, x = arcsin (- \frac{- 1 - 1 + 16 q ( - a + 2 q )}{8 q}) + 2 πn, x = π + arcsin (\frac{- 1 - 1 + 16 q ( - a + 2 q )}{8 q}) + 2 πn

Solution steps

sin (x) = a - 2 q cos (2 x)

Subtract

a - 2 q cos (2 x)

from both sides

sin (x) - a + 2 q cos (2 x) = 0

Rewrite using trig identities

sin (x) - a + (1 - 2 sin^{2} (x)) \cdot 2 q = 0

Solve by substitution

sin (x) = - \frac{- 1 + 1 + 16 q ( - a + 2 q )}{8 q}, sin (x) = - \frac{- 1 - 1 + 16 q ( - a + 2 q )}{8 q}; q \neq = 0

sin (x) = - \frac{- 1 + 1 + 16 q ( - a + 2 q )}{8 q} : x = arcsin (- \frac{- 1 + 1 + 16 q ( - a + 2 q )}{8 q}) + 2 πn, x = π + arcsin (\frac{- 1 + 1 + 16 q ( - a + 2 q )}{8 q}) + 2 πn

sin (x) = - \frac{- 1 - 1 + 16 q ( - a + 2 q )}{8 q} : x = arcsin (- \frac{- 1 - 1 + 16 q ( - a + 2 q )}{8 q}) + 2 πn, x = π + arcsin (\frac{- 1 - 1 + 16 q ( - a + 2 q )}{8 q}) + 2 πn

Combine all the solutions

x = arcsin (- \frac{- 1 + 1 + 16 q ( - a + 2 q )}{8 q}) + 2 πn, x = π + arcsin (\frac{- 1 + 1 + 16 q ( - a + 2 q )}{8 q}) + 2 πn, x = arcsin (- \frac{- 1 - 1 + 16 q ( - a + 2 q )}{8 q}) + 2 πn, x = π + arcsin (\frac{- 1 - 1 + 16 q ( - a + 2 q )}{8 q}) + 2 πn

cos (2 x) - 3 sin (- x) + 2 = 0

0 = - sin (8 x) + sin (6 x)

3 tan (x) sin (x) - 3 sin (x) = 0

cot^{4} (x) - 3 cot^{2} (x) + 1 = 0

\frac{cos ( - 3 0 ^{\circ} + x )}{cos ( 3 0 ^{\circ} + x )} = \frac{1835}{726}

What is the general solution for solvefor x,sin(x)=a-2*q*cos(2x) ?
The general solution for solvefor x,sin(x)=a-2*q*cos(2x) is