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

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

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Solve over the Interval sin(2x)-2cos(x)=0 , (0,2pi)

sin (2 x) - 2 cos (x) = 0, (0, 2 π)

x = \frac{π}{2}, x = \frac{3 π}{2}

Solution steps

sin (2 x) - 2 cos (x) = 0, (0, 2 π)

Rewrite using trig identities

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

Factor

- 2 cos (x) + 2 cos (x) sin (x) : 2 cos (x) (sin (x) - 1)

2 cos (x) (sin (x) - 1) = 0

Solving each part separately

cos (x) = 0 or sin (x) - 1 = 0

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

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

Combine all the solutions

x = \frac{π}{2}, x = \frac{3 π}{2}

sin (\frac{( 5 π )}{3})

sin (\frac{( 5 π )}{2})

sin (\frac{( 3 π )}{2})

x, k = \frac{( 1 + sin ( x ) )}{n}

a, tan (a) = \frac{1}{3}

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