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Chapter 29: Chapter 29

Problem 925

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Solve \(2 \sin ^{2} \theta+3 \cos \theta-3=0\) for \(\theta\) if $0 \leq \theta<360^{\circ}$.

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The possible values for \(\theta\) that satisfy the given equation are: \(\theta = 0^{\circ}, 60^{\circ}, 300^{\circ}\).
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TABLE OF CONTENTS :
TABLE OF CONTENTS

Step 1: Use the identity \(\sin^2 \theta = 1 - \cos^2 \theta\)

We are given the equation \(2\sin^2 \theta + 3\cos\theta - 3 = 0\). To convert this equation to a single trigonometric function, we can use the identity \(\sin^2 \theta = 1 - \cos^2 \theta\). Substituting this identity into the equation, we get: $$2(1 - \cos^2 \theta) + 3\cos\theta - 3 = 0.$$

Step 2: Simplify the equation

Now, we need to simplify the equation. Expanding and simplifying, we have: $$2 - 2\cos^2 \theta + 3\cos\theta - 3 = 0.$$ This can be written as a quadratic equation in terms of the cosine function: $$2\cos^2 \theta - 3\cos\theta + 1 = 0.$$

Step 3: Solve the quadratic equation

To find the values of \(\cos\theta\), we need to solve the quadratic equation \(2\cos^2 \theta - 3\cos\theta + 1 = 0\). Let's consider \(x = \cos \theta\). The equation can be written as: $$2x^2 - 3x + 1 = 0.$$ By factoring the quadratic equation, we get: $$(2x - 1)(x - 1) = 0.$$ This equation has two possible solutions: $$x_1 = \frac{1}{2}, \quad x_2 = 1.$$

Step 4: Find the values of \(\theta\)

Since \(x = \cos \theta\), we have two possible values for the cosine function: $$\cos \theta_1 = \frac{1}{2}, \quad \cos \theta_2 = 1.$$ We need to find the angles \(\theta\) that satisfy these values of cosine function within the domain \(0 \leq \theta < 360^{\circ}\). For \(\cos \theta_1 = \frac{1}{2}\): $$\theta_1 = 60^{\circ}, \quad \theta_2 = 300^{\circ}.$$ For \(\cos \theta_2 = 1\): $$\theta_3 = 0^{\circ}.$$ Therefore, the possible values for \(\theta\) that satisfy the given equation are: $$\theta = 0^{\circ}, 60^{\circ}, 300^{\circ}.$$

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