What is the general solution for cos(2x)-sin(-x)=0 ?

The general solution for cos(2x)-sin(-x)=0 is x=(7pi)/6+2pin,x=(11pi)/6+2pin,x= pi/2+2pin

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

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

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

Solution steps

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

sin (x) = 1 : x = \frac{π}{2} + 2 πn

Combine all the solutions

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

sin (x) cos (x) = \frac{1}{2} cos (x)

sin (θ) - 2 sin (θ) cos (θ) = 0

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

sin (2 π t) = 1

(4 cos^{2} (x) - 3) (csc (x) + 2) = 0

What is the general solution for cos(2x)-sin(-x)=0 ?
The general solution for cos(2x)-sin(-x)=0 is x=(7pi)/6+2pin,x=(11pi)/6+2pin,x= pi/2+2pin