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I am required to write the equation $$\cot x - \cos x = 0$$ in the form $$\cos x(1 - \sin x) = 0$$

What I reached is as follows, \begin{align} \cot x & = \frac{\cos x}{\sin x}\\[4pt] \frac{\cos x}{\sin x} - \cos x & = 0\\[4pt] \cos x\left(\frac{1}{\sin x} - 1\right) & = 0\\[4pt] \cos x\left(\frac{1}{\sin x} - \frac{\sin x}{\sin x}\right) & = 0 \end{align}

How can I rewrite in the above format?

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  • $\begingroup$ You're almost done. Under the assumption that $\sin x \neq 0$, you can remove it from the denominator to obtain $\cos X (1-\sin X) = \sin X \cdot 0 = 0$. If it were zero, then $\cot X = \infty$, so the statement is not possible. $\endgroup$ – астон вілла олоф мэллбэрг Sep 10 '16 at 10:31
  • $\begingroup$ How exactly can I remove the sinX $\endgroup$ – user367737 Sep 10 '16 at 10:31
  • $\begingroup$ Take it to the other side, to be multiplied by zero, that's how it's removed. Should I answer the question more clearly. $\endgroup$ – астон вілла олоф мэллбэрг Sep 10 '16 at 10:32
  • $\begingroup$ Oh I get it thank you very much $\endgroup$ – user367737 Sep 10 '16 at 10:33
  • $\begingroup$ I wrote an answer anyway. I want to make sure you are satisfied. $\endgroup$ – астон вілла олоф мэллбэрг Sep 10 '16 at 10:36
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$$ \cot(X)-\cos(X) = 0 \implies \frac{\cos X}{\sin X} - \cos X = 0 \implies \cos X\bigg(\frac{1}{\sin X} - 1\bigg) = 0 $$ $$ \implies \frac{\cos X(1 - \sin X)}{\sin X} = 0 \implies \cos X(1 - \sin X) = 0 \cdot \sin X = 0 $$

This applies when $\sin X \neq 0$. If it is zero, then $\cot X = \infty$ so the equation is not satisfied.

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