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

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

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

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

x = \frac{π}{2}, x = \frac{3 π}{2}

Solution steps

sin (2 x) + 2 cos (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}

cos (x) = \frac{8.5}{9.9}

cos^{3} (x) = - cos^{2} (x)

3 tan (2 x - π) + 1 = 0

3 sec (2 x) + 2 = 0

e^{t a n (x)} - 2 = 0

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