What is the general solution for sin(2x)=cos(x),0<= x<2pi ?

The general solution for sin(2x)=cos(x),0<= x<2pi is x= pi/2 ,x=(3pi)/2 ,x= pi/6 ,x=(5pi)/6

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sin(2x)=cos(x),0<= x<2pi

sin (2 x) = cos (x), 0 \leq x < 2 π

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

Solution steps

sin (2 x) = cos (x), 0 \leq x < 2 π

Subtract

cos (x)

from both sides

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

Rewrite using trig identities

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

Factor

- cos (x) + 2 cos (x) sin (x) : cos (x) (2 sin (x) - 1)

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

Solving each part separately

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

cos (x) = 0, 0 \leq x < 2 π : x = \frac{π}{2}, x = \frac{3 π}{2}

2 sin (x) - 1 = 0, 0 \leq x < 2 π : x = \frac{π}{6}, x = \frac{5 π}{6}

Combine all the solutions

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

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

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

2 \cdot cos (\frac{x}{2}) - cos (x) = 1

2 sin (θ) cos (θ) + 2 sin (θ) - 2 cos (θ) sin (θ) = 0

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

What is the general solution for sin(2x)=cos(x),0<= x<2pi ?
The general solution for sin(2x)=cos(x),0<= x<2pi is x= pi/2 ,x=(3pi)/2 ,x= pi/6 ,x=(5pi)/6