Question

a) Helene is asked to solve the equation \(3 \sin ^{2} \theta-2 \sin \theta=0,0 \leq \theta \leq \pi .\) She finds that \(\theta=\pi .\) Show how she could check whether this is a correct root for the equation. b) Find all the roots of the equation \(3 \sin ^{2} \theta-2 \sin \theta=0, \theta \in[0, \pi]\)

Short answer

The roots of the equation are \( \theta = 0, \arcsin \left( \frac{2}{3} \right), \pi \).

Step-by-step solution

Rewrite the original equation

Start with the given equation: \[ 3 \, \sin^{2} \theta - 2 \, \sin \theta = 0 \]

Factor the equation

Factor out the common term \( \sin \theta \) from the left side of the equation: \[ \sin \theta (3 \, \sin \theta - 2) = 0 \]

Set each factor to zero and solve

Set each factor equal to zero and solve for \( \theta \):\[ \sin \theta = 0 \quad \text{or} \quad 3 \, \sin \theta - 2 = 0 \]For \( \sin \theta = 0 \):\[ \theta = 0, \, \pi \]For \( 3 \, \sin \theta - 2 = 0 \):\[ 3 \sin \theta = 2 \ \sin \theta = \frac{2}{3} \]

Find the angles that satisfy the equation \( \sin \theta = \frac{2}{3} \)

Since \( \theta \) is in the range \( [0, \pi] \), find the angle \( \theta \) such that \( \sin \theta = \frac{2}{3} \) in this range. Using the inverse sine function, \( \theta = \arcsin \left( \frac{2}{3} \right) \).

Combine all solutions

The solutions obtained are \( \theta = 0, \arcsin \left( \frac{2}{3} \right), \pi \). Collect all the valid solutions within the given range \( 0 \leq \theta \leq \pi \).

Check if \( \theta = \pi \) is a root (Part a)

Substitute \( \theta = \pi \) back into the original equation to verify it is a root: \[ 3 \sin^{2}(\pi) - 2 \sin(\pi) \ = 3 (0)^{2} - 2 (0) \ = 0 \]Therefore, \( \theta = \pi \) is a correct root.

Key concepts

factoring trigonometric expressions

Factoring trigonometric expressions is a key skill to master in trigonometry. When you approach an equation like the one Helene encountered, \(3 \sin^{2} \theta - 2 \sin \theta = 0\), your first step should be to simplify it by factoring.

Look for common terms in the equation. In this case, \(\sin \theta\) appears in both terms. Factoring out \(\sin \theta\) from the left side gives: \[ \sin \theta (3 \sin \theta - 2) = 0 \]

The equation is now in a form where we can apply the zero-product property. This property states that if the product of two factors is zero, then at least one of the factors must be zero. This leads us to the next crucial step: setting each factor to zero separately:

• \(\sin \theta = 0\)
• \(3 \sin \theta - 2 = 0\)

Solving these equations individually will give us potential solutions for \(\theta\). This method simplifies the process by breaking it down into more manageable parts.

inverse sine function

The inverse sine function, also known as \(\text{arcsin}\), is useful for finding the angle that corresponds to a given sine value. In trigonometric equation solving, it's particularly handy when your equation yields a sine value that isn't straightforward.

For example, from the equation \(3 \sin \theta - 2 = 0\), you get: \[ 3 \sin \theta = 2 \sin \theta = \frac{2}{3} \]

Now, we need to find the angle \(\theta\) for which \(\sin \theta = \frac{2}{3}\). This is where the inverse sine function comes into play.

Applying the inverse sine gives us: \[ \theta = \text{arcsin}\big(\frac{2}{3}\big) \]

Given that \(\theta\)'s value must lie within the range \(0 \leq \theta \leq \pi\), using \(\text{arcsin}\) ensures that the solution we obtain falls within this interval. Always remember to consider the function's domain and range when using inverse trigonometric functions.

verifying trigonometric solutions

Verifying trigonometric solutions helps ensure our answers are correct and fall within the specified range. For Helene's problem, once she found \(\theta = \pi\) as a solution, she needed to check if it was indeed correct.

To verify \(\theta = \pi\), substitute this value back into the original equation: \[ 3 \sin^{2}(\theta) - 2 \sin(\theta) \]

For \(\theta = \pi\), we have \(\sin(\pi) = 0\). Plugging this into the equation gives: \[ 3 \sin^{2}(\pi) - 2 \sin(\pi) \]

This simplifies to: \[ 3 (0)^{2} - 2 (0) = 0 \]

The left-hand side and the right-hand side are equal, confirming that \(\theta = \pi\) is indeed a valid solution.

It's important to verify each solution this way to avoid errors. This step ensures you've found all the correct solutions and confirms their accuracy within the given domain.