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Problem 925

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

Expert verified
The possible values for $$\theta$$ that satisfy the given equation are: $$\theta = 0^{\circ}, 60^{\circ}, 300^{\circ}$$.
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## 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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