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

The general solution for tan(2θ)+2cos(θ)=0,0<= θ<= 2pi is θ= pi/2 ,θ=(3pi)/2 ,θ=(7pi)/6 ,θ=(11pi)/6

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tan(2θ)+2cos(θ)=0,0<= θ<= 2pi

tan (2 θ) + 2 cos (θ) = 0, 0 \leq θ \leq 2 π

θ = \frac{π}{2}, θ = \frac{3 π}{2}, θ = \frac{7 π}{6}, θ = \frac{11 π}{6}

Solution steps

tan (2 θ) + 2 cos (θ) = 0, 0 \leq θ \leq 2 π

Express with sin, cos

\frac{sin ( 2 θ ) + 2 cos ( 2 θ ) cos ( θ )}{cos ( 2 θ )} = 0

\frac{f ( x )}{g ( x )} = 0 \Rightarrow f (x) = 0

sin (2 θ) + 2 cos (2 θ) cos (θ) = 0

Rewrite using trig identities

2 cos (2 θ) cos (θ) + 2 cos (θ) sin (θ) = 0

Factor

2 cos (2 θ) cos (θ) + 2 cos (θ) sin (θ) : 2 cos (θ) (cos (2 θ) + sin (θ))

2 cos (θ) (cos (2 θ) + sin (θ)) = 0

Solving each part separately

cos (θ) = 0 or cos (2 θ) + sin (θ) = 0

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

cos (2 θ) + sin (θ) = 0, 0 \leq θ \leq 2 π : θ = \frac{7 π}{6}, θ = \frac{11 π}{6}, θ = \frac{π}{2}

Combine all the solutions

θ = \frac{π}{2}, θ = \frac{3 π}{2}, θ = \frac{7 π}{6}, θ = \frac{11 π}{6}

cos (2 x) + 10 sin^{2} (x) = 5

2 = 3 + 2 sin (θ)

- 2 sin (θ) + sin (2 θ) = 0

- 2 sin^{2} (x) - 3 sin (x) + 2 = 0

tan (β) = \frac{1}{2}

What is the general solution for tan(2θ)+2cos(θ)=0,0<= θ<= 2pi ?
The general solution for tan(2θ)+2cos(θ)=0,0<= θ<= 2pi is θ= pi/2 ,θ=(3pi)/2 ,θ=(7pi)/6 ,θ=(11pi)/6