Solve the equation for exact solutions over the interval $[0, 2 π)$ .
$3 \cot x + 2 = 5$
Select the correct choice below and, if necessary, fill in the answer box to complete your choice.
A. The solution set is
(Type an exact answer, using $π$ as needed. Type your answer in radians. Use integers or fractions for any numbers in the expression. Use a comma to separate answers as needed.)
B. The solution is the empty set

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However, if $\sin x = - \frac{\sqrt{2}}{2},$ then $\cos x = - \frac{\sqrt{2}}{2},$ which means $\cot x$ is negative. So $\sin x = \frac{\sqrt{2}}{2},$ which leads to the $2$ solutions $x = \frac{π}{4}$ and $x = \frac{7 π}{4} .$ We check that both solutions work.

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Step 1 ：We can write the equation as $3 \frac{\cos x}{\sin x} + 2 = 5.$

Step 2 ：Then $3 \frac{\cos x}{\sin x} = 3,$ so $\frac{\cos x}{\sin x} = 1.$

Step 3 ：This implies $\cos x = \sin x .$

Step 4 ：We know that $\sin^{2} x + \cos^{2} x = 1,$ so $2 \sin^{2} x = 1,$ which gives $\sin^{2} x = \frac{1}{2} .$

Step 5 ：This equation factors as $(\sin x - \frac{\sqrt{2}}{2}) (\sin x + \frac{\sqrt{2}}{2}) = 0,$ so $\sin x = \frac{\sqrt{2}}{2}$ or $\sin x = - \frac{\sqrt{2}}{2} .$

Step 6 ：However, if $\sin x = - \frac{\sqrt{2}}{2},$ then $\cos x = - \frac{\sqrt{2}}{2},$ which means $\cot x$ is negative. So $\sin x = \frac{\sqrt{2}}{2},$ which leads to the $2$ solutions $x = \frac{π}{4}$ and $x = \frac{7 π}{4} .$ We check that both solutions work.

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