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

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

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

cos (x) = sin (2 x), 0 \leq x \leq 2 π

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

Solution steps

cos (x) = sin (2 x), 0 \leq x \leq 2 π

Subtract

sin (2 x)

from both sides

cos (x) - sin (2 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 \leq 2 π : x = \frac{π}{2}, x = \frac{3 π}{2}

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

tan (θ) (sin (θ) - 1) = 0

sin (θ) - 4 = - 5

sin (θ) - 4 = - 3

csc^{2} (x) = 4 cot^{2} (x)

cot^{2} (3 x) - tan^{2} (3 x) = 0

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