Question

Solve $\sin^2x = 0$ over the domain $0^\circ \le x \le 360^\circ$. Explain/show all steps to reach your answer.

          Solve $\sin^2x = 0$ over the domain $0^\circ \le x \le 360^\circ$. Explain/show all steps to reach your answer.
        
Solve sin^2x = 0 over the domain 0^∘≤ x ≤ 360^∘. Explain/show all steps to reach your answer.

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Precalculus with Limits
Precalculus with Limits
Ron Larson 2nd Edition
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Solve sin^(2)x=0 over the domain 0deg <=x<=360deg . Explain/show all steps to reach your answer. Solve sinx=I over the domain x360.Explain/show all steps to reach your answer
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Transcript

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00:01 2 divided by x squared minus x plus 1 divided by x plus 3 equal to 4 divided by x minus 2 or 2 divided by x into x minus 1 plus 1 divided by x plus 1 divided by x plus 3 so here either x is not equal to 0 or x is not equal to 1.
00:49 So the domains of the three expressions are set of x such that x is a real number whereas x is not to 0 and x is not equal to 1 next set of x such that x is a real number whereas x is not equal to minus 3 and set of x such that x is a real number whereas x is not equal to 2...
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