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

The general solution for sin(2x)=2cos^2(x),0<= x<= 2pi is x= pi/2 ,x=(3pi)/2 ,x= pi/4 ,x=(5pi)/4

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

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

x = \frac{π}{2}, x = \frac{3 π}{2}, x = \frac{π}{4}, x = \frac{5 π}{4}

Solution steps

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

Subtract

2 cos^{2} (x)

from both sides

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

Rewrite using trig identities

- 2 cos^{2} (x) + 2 cos (x) sin (x) = 0

Factor

- 2 cos^{2} (x) + 2 cos (x) sin (x) : 2 cos (x) (- cos (x) + sin (x))

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

Solving each part separately

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

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

- cos (x) + sin (x) = 0, 0 \leq x \leq 2 π : x = \frac{π}{4}, x = \frac{5 π}{4}

Combine all the solutions

x = \frac{π}{2}, x = \frac{3 π}{2}, x = \frac{π}{4}, x = \frac{5 π}{4}

2 sin (x) + 1 = cos (2 x)

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

cos (x) = \frac{1}{6}

2 cos^{2} (θ) + 9 cos (θ) + 4 = 0

3 sin^{2} (x) - 5 sin (x) - 2 = 0

What is the general solution for sin(2x)=2cos^2(x),0<= x<= 2pi ?
The general solution for sin(2x)=2cos^2(x),0<= x<= 2pi is x= pi/2 ,x=(3pi)/2 ,x= pi/4 ,x=(5pi)/4