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Popular >

3sin(θ)+2sin^2(θ)=-1,0<= θ<2pi

Solution

3 sin (θ) + 2 sin^{2} (θ) = - 1, 0 \leq θ < 2 π

Solution

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

+1

Solution steps

3 sin (θ) + 2 sin^{2} (θ) = - 1, 0 \leq θ < 2 π

Solve by substitution

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

sin (θ) = - \frac{1}{2}, 0 \leq θ < 2 π : θ = \frac{7 π}{6}, θ = \frac{11 π}{6}

sin (θ) = - 1, 0 \leq θ < 2 π : θ = \frac{3 π}{2}

Combine all the solutions

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

Popular Examples

3sin^2(x)+4sin(x)+1=0

3 sin^{2} (x) + 4 sin (x) + 1 = 0

cos (x) = \frac{9}{14}

sin (θ) = \frac{7}{9}

cos(θ)=0.97617

cos (θ) = 0.97617

cos(a)=0,999^{99}x,(sin(a))/(cos(a))x

cos (a) = 0, 99 9^{99} x, \frac{sin ( a )}{cos ( a )} x

Frequently Asked Questions (FAQ)

What is the general solution for 3sin(θ)+2sin^2(θ)=-1,0<= θ<2pi ?
The general solution for 3sin(θ)+2sin^2(θ)=-1,0<= θ<2pi is θ=(7pi)/6 ,θ=(11pi)/6 ,θ=(3pi)/2