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

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

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

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

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

Solution steps

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

Rewrite using trig identities

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

Factor

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

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

Solving each part separately

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

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

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

Combine all the solutions

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

csc (θ) = - \frac{2}{3}

2 cos (\frac{x}{2}) + 3 = 0

\frac{1}{cos ( x )} = - 2

25 sin (x) cos (x) - 5 sin (x) + 20 cos (x) = 4

sin (θ) = - 0.9

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