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

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

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solvefor x,sin(x)=cos(2x)

solve for

x, sin (x) = 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

sin (x) = cos (2 x)

Subtract

cos (2 x)

from both sides

sin (x) - 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

cos (4 t) = 0

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

8 cos^{2} (θ) - 3 cos (θ) = 0, 0^{\circ} \leq θ \leq 18 0^{\circ}

8 sin (x) - 7 = 0

1 - sin (θ) = 1 + cos (θ)

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