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

The general solution for 2sin^2(θ)+cos(θ)=0,0<= θ<= 2pi is θ=2.46670…,θ=-2.46670…+2pi

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2sin^2(θ)+cos(θ)=0,0<= θ<= 2pi

2 sin^{2} (θ) + cos (θ) = 0, 0 \leq θ \leq 2 π

θ = 2.46670 \dots, θ = - 2.46670 \dots + 2 π

Solution steps

2 sin^{2} (θ) + cos (θ) = 0, 0 \leq θ \leq 2 π

Rewrite using trig identities

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

Solve by substitution

cos (θ) = - \frac{- 1 + 17}{4}, cos (θ) = \frac{1 + 17}{4}

cos (θ) = - \frac{- 1 + 17}{4}, 0 \leq θ \leq 2 π : θ = arccos (- \frac{17 - 1}{4}), θ = - arccos (- \frac{- 1 + 17}{4}) + 2 π

cos (θ) = \frac{1 + 17}{4}, 0 \leq θ \leq 2 π :

No Solution

Combine all the solutions

θ = arccos (- \frac{17 - 1}{4}), θ = - arccos (- \frac{- 1 + 17}{4}) + 2 π

Show solutions in decimal form

θ = 2.46670 \dots, θ = - 2.46670 \dots + 2 π

cos (a) - 1 = 2 cos (a)

3 cos^{2} (θ) - sin (θ) = 1

5 cos (2 x) = 9 sin (x) + 4

- 3 cos (x) - 1 = cos (2 x)

- 2 tan^{2} (θ) - 4 tan (θ) + 2 = 3 tan (θ) + 5

What is the general solution for 2sin^2(θ)+cos(θ)=0,0<= θ<= 2pi ?
The general solution for 2sin^2(θ)+cos(θ)=0,0<= θ<= 2pi is θ=2.46670…,θ=-2.46670…+2pi