# Find all solutions of the equation. 3\sin^{2}(\theta) - 7\sin(\theta) + 2 = 0

## Question:

Find all solutions of the equation.

{eq}\displaystyle\; 3\sin^{2}(\theta) - 7\sin(\theta) + 2 = 0 {/eq}

## Trigonometric Equations:

Trigonometric equations are equations made of trigonometric functions. For example, {eq}\text{sine(sin), tangent(tan), cosine(cos)} {/eq}. If a trigonometric equation holds for all angles, then it is a trigonometric identity. For example, {eq}\sin^2x + \cos^2 x = 1 {/eq} is a trigonometric equation which is an identity.

## Answer and Explanation:

We have the trigonometric equation:

$$3\sin^{2}(\theta) - 7\sin(\theta) + 2 = 0 $$

Let {eq}\sin (\theta) = x {/eq}. Thus, we can rewrite the equation as:

$$3x^2 - 7x + 2 = 0 $$

Factoring the left-hand side of the equation, we get:

$$(x - 2)(3x - 1) = 0 $$

Setting each factor equal to zero:

$$\begin{align} x - 2 &= 0 \\[0.3cm] x &= 2 & \left( \text{i}\right) \end{align} $$

Also,

$$\begin{align} (3x - 1) &= 0 \\[0.3cm] x &= \dfrac{1}{3} & \left( \text{ii}\right) \end{align} \\ $$

From the first solution, we have:

$$x = \sin \theta = 2 $$

Since the range of sine is between {eq}-1{/eq} and {eq}1 {/eq}, there is no solution for this.

From solution {eq}\text{(ii)} {/eq}, we have:

$$\begin{align} x = \sin \theta &= \dfrac{1}{3} \\[0.3cm] \end{align} $$

Taking the inverse of sine on both sides of the equation, we get:

$$\theta =\sin^{-1} \left(\dfrac{1}{3} \right) \approx 19.4712^\circ $$

The value of sine is positive in the first and in the second quadrant. Therefore:

$$\theta = 180 - 19.4712 = 160.5288^\circ $$

Since the period of sine is {eq}360^\circ \text{ or } 2\pi {/eq}, the solutions will repeat every {eq}360^\circ {/eq} on either directions. Therefore:

$$\boxed{\color{blue}{\theta = (19.4712 + 360k, \quad 160.5288 + 360k)}} $$

Where {eq}k {/eq} is any integer.

#### Ask a question

Our experts can answer your tough homework and study questions.

Ask a question Ask a question#### Search Answers

#### Learn more about this topic:

from High School Precalculus Textbook

Chapter 22 / Lesson 5