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

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

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

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

x = 0, x = \frac{2 π}{3}, x = \frac{4 π}{3}, x = 2 π

Solution steps

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

Rewrite using trig identities

cos (x) = cos (2 x)

Subtract

cos (2 x)

from both sides

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

Rewrite using trig identities

2 sin (\frac{3 x}{2}) sin (\frac{x}{2}) = 0

Solving each part separately

sin (\frac{3 x}{2}) = 0 or sin (\frac{x}{2}) = 0

sin (\frac{3 x}{2}) = 0, 0 \leq x \leq 2 π : x = 0, x = \frac{2 π}{3}, x = \frac{4 π}{3}, x = 2 π

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

Combine all the solutions

x = 0, x = \frac{2 π}{3}, x = \frac{4 π}{3}, x = 2 π

cos (x) = - 0.397

sin (2 x) cos (x) + cos (2 x) sin (x) = \frac{3}{2}

cos (x) = \frac{5 2}{10}

4 tan (x) + 4 = 0

4 sin (θ) - 3 = 0, 0 \leq θ \leq 2 π

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