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

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

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

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

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

Solution steps

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

Subtract

sin (x)

from both sides

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

Rewrite using trig identities

1 - sin (x) - 2 sin^{2} (x) = 0

Solve by substitution

sin (x) = - 1, sin (x) = \frac{1}{2}

sin (x) = - 1, 0 \leq x \leq 2 π : x = \frac{3 π}{2}

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

Combine all the solutions

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

tan (x - \frac{π}{2}) = 1

5 cos (x) = 4 sin (x) + 4

6 sec^{2} (x) tan (x) = 8 tan (x)

csc (θ) = 0

- sin^{2} (θ) + cos^{2} (θ) = - sin (θ) + 1

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