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

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

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

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

x = \frac{π}{2}, x = \frac{3 π}{2}

Solution steps

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

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

Combine all the solutions

x = \frac{π}{2}, x = \frac{3 π}{2}

tan^{2} (x) sin (x) = sin (x)

2 sin (2 x) - 3 = 0

2 sin (x) cos (x) = 0

tan^{5} (x) - 9 tan (x) = 0

1 - cos (θ) = \frac{1}{2}

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