How do you solve $2 {cos}^{2} θ + 3 sin θ = 0$ $2cos^2theta+3sintheta=0$ ?

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How do you solve $2 {cos}^{2} θ + 3 sin θ = 0$ $2cos^2theta+3sintheta=0$ ?

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Answer 1 · 2016-11-12T22:10:14

$2 (1 - {sin}^{2} θ) + 3 sin θ = 0$ $2(1- sin^2theta) + 3sintheta= 0$

$2 - 2 {sin}^{2} θ + 3 sin θ = 0$ $2 - 2sin^2theta + 3sintheta = 0$

$0 = 2 {sin}^{2} θ - 3 sin θ - 2$ $0= 2sin^2theta - 3sintheta - 2$

$0 = 2 {sin}^{2} θ - 4 sin θ + sin θ - 2$ $0 = 2sin^2theta - 4sintheta + sin theta - 2$

$0 = 2 sin θ (sin θ - 2) + 1 (sin θ - 2)$ $0 = 2sintheta(sin theta - 2) + 1(sin theta -2)$

$0 = (2 sin θ + 1) (sin θ - 2)$ $0 = (2sintheta + 1)(sin theta - 2)$

$sin θ = - \frac{1}{2} and sin θ = 2$ $sintheta= -1/2 and sintheta= 2$

$θ = \frac{7 π}{6}, \frac{11 π}{6}$ $theta= (7pi)/6, (11pi)/6$

Hopefully this helps!

Answer 2 · 2016-11-12T22:12:00

Please see the explanation.

Explanation:

Substitute $1 - {sin}^{2} (θ)$ $1 - sin^2(theta)$ for ${cos}^{2} (θ)$ $cos^2(theta)$ :

$2 (1 - {sin}^{2} (θ)) + 3 sin (θ) = 0$ $2(1 - sin^2(theta)) + 3sin(theta) = 0$

Use the distributive property:

$2 - 2 {sin}^{2} (θ)) + 3 sin (θ) = 0$ $2 - 2sin^2(theta)) + 3sin(theta) = 0$

Multiply both side by -1:

$2 {sin}^{2} (θ)) - 3 sin (θ) - 2 = 0$ $2sin^2(theta)) - 3sin(theta) - 2 = 0$

This is a quadratic where the variable is $sin (θ)$ $sin(theta)$ .

It looks like it will factor:

#(sin(theta) - 2)(2sin(theta) + 1) = 0

$sin (θ) = 2 and sin (θ) = - \frac{1}{2}$ $sin(theta) = 2 and sin(theta) = -1/2$

We must discard the first root, because it is outside the range of the sine function.

Turning our attention to the second root:

$sin (θ) = - \frac{1}{2}$ $sin(theta) = -1/2$

Rotating counterclockwise from 0, the first encounter of this is at:

$θ = \frac{7 π}{6}$ $theta = (7pi)/6$

The next encounter with this is at:

$θ = \frac{11 π}{6}$ $theta = (11pi)/6$

Add integer rotations of $2 π$ $2pi$ to both:

$θ = \frac{7 π}{6} + 2 n π$ $theta = (7pi)/6 + 2npi$ and $θ = \frac{11 π}{6} + 2 n π$ $theta = (11pi)/6 + 2npi$ Where n can be any integer (positive, negative, or zero)

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