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Problem 926

Find the solution set of $$\sin ^{2} \theta+\sin \theta=0$$.

Expert verified
The solution set of $$\sin ^{2} \theta+\sin \theta=0$$ is given by $$\theta = 0 + n\pi$$, where n is any integer or $$\theta = (-1)^{n}\pi$$, where n is any odd integer.
See the step by step solution

Step 1: Factor the given equation

When we analyze the given equation, we can factor out a common term like this: $\sin\theta( \sin\theta + 1) = 0.$

Step 2: Equate the factors to zero

Now, we have two factors that multiply to make zero. This means that at least one of the factors should be equal to zero. Thus, in solving this equation, we must consider both the possibilities: $\sin\theta = 0$ or $\sin\theta + 1 = 0.$

Step 3: Solve for θ

Now, we solve each equation for $$\theta$$: For the first equation: $$\sin\theta = 0$$, we get $\theta = 0 + n\pi,$ where n is any integer. For the second equation: $$\sin\theta + 1 = 0$$, we get $\sin\theta = -1,$ which has the general solution $\theta = (-1)^{n}\pi,$ where n is any odd integer.

Step 4: Combine both sets of solutions

Finally, we can combine both families of solutions into a single set: $\theta = 0 + n\pi,$ where n is any integer or $\theta = (-1)^{n}\pi,$ where n is any odd integer. Now we've found the set of solutions that fulfill the given condition in the exercise. This set contains the general solutions for $$\theta$$.

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