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

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

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

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

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

Solution steps

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

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}, 0 \leq x \leq 2 π : x = \frac{π}{6}, x = \frac{5 π}{6}

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

Combine all the solutions

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

sin (2 x) = sin (x) cos (x)

sin (θ) = - \frac{5}{5}, \frac{3 π}{2} < θ < 2 π

(2 sin (2 x) - 1) (3 tan (2 x) - 1) = 0

- 0.2588 + 0.9659 tan (b) = 1.4201

2 cot^{2} (t) sin (t) - cot^{2} (t) = 0

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