What is the general solution for -4cos(θ)=-sin^2(θ)+1 ?

The general solution for -4cos(θ)=-sin^2(θ)+1 is θ= pi/2+2pin,θ=(3pi)/2+2pin

Popular >

-4cos(θ)=-sin^2(θ)+1

- 4 cos (θ) = - sin^{2} (θ) + 1

θ = \frac{π}{2} + 2 πn, θ = \frac{3 π}{2} + 2 πn

Solution steps

- 4 cos (θ) = - sin^{2} (θ) + 1

Subtract

- sin^{2} (θ) + 1

from both sides

- 4 cos (θ) + sin^{2} (θ) - 1 = 0

Rewrite using trig identities

- cos^{2} (θ) - 4 cos (θ) = 0

Solve by substitution

cos (θ) = - 4, cos (θ) = 0

cos (θ) = - 4 :

No Solution

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

Combine all the solutions

θ = \frac{π}{2} + 2 πn, θ = \frac{3 π}{2} + 2 πn

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

sin (θ) = \frac{π}{2}

arcsin (\frac{900 x ^{2} - 1}{900 x ^{2} + 1}) = 1.18

sin (b - 2 0^{\circ}) = cos (142. 7^{\circ})

2 \cdot (1 - cos^{2} (x)) - cos^{2} (x) = 1 + sin^{2} (x)

What is the general solution for -4cos(θ)=-sin^2(θ)+1 ?
The general solution for -4cos(θ)=-sin^2(θ)+1 is θ= pi/2+2pin,θ=(3pi)/2+2pin