What is the general solution for 3sin^2(2θ)+8cos(2θ)=0 ?

The general solution for 3sin^2(2θ)+8cos(2θ)=0 is θ=(1.91063…)/2+pin,θ=-(1.91063…)/2+pin

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3sin^2(2θ)+8cos(2θ)=0

3 sin^{2} (2 θ) + 8 cos (2 θ) = 0

θ = \frac{1.91063 \dots}{2} + πn, θ = - \frac{1.91063 \dots}{2} + πn

Solution steps

3 sin^{2} (2 θ) + 8 cos (2 θ) = 0

Rewrite using trig identities

(1 - cos^{2} (2 θ)) \cdot 3 + 8 cos (2 θ) = 0

Solve by substitution

cos (2 θ) = - \frac{1}{3}, cos (2 θ) = 3

cos (2 θ) = - \frac{1}{3} : θ = \frac{arccos ( - \frac{1}{3} )}{2} + πn, θ = - \frac{arccos ( - \frac{1}{3} )}{2} + πn

cos (2 θ) = 3 :

No Solution

Combine all the solutions

θ = \frac{arccos ( - \frac{1}{3} )}{2} + πn, θ = - \frac{arccos ( - \frac{1}{3} )}{2} + πn

Show solutions in decimal form

θ = \frac{1.91063 \dots}{2} + πn, θ = - \frac{1.91063 \dots}{2} + πn

cos (2 x - \frac{π}{3}) = \frac{1}{2}

sin (θ) = \frac{7}{11}

12 sin (t) cos (t) = 8 sin (t)

4 sin^{2} (x) - 8 sin (x) + 3 = 0

3 tan (2 x) + 3 = 0

What is the general solution for 3sin^2(2θ)+8cos(2θ)=0 ?
The general solution for 3sin^2(2θ)+8cos(2θ)=0 is θ=(1.91063…)/2+pin,θ=-(1.91063…)/2+pin