# Transforming the equation $\cot x -\cos x = 0$ into the form $\cos x(1- \sin x) = 0$

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?

• 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. – астон вілла олоф мэллбэрг Sep 10 '16 at 10:31
• How exactly can I remove the sinX – user367737 Sep 10 '16 at 10:31
• Take it to the other side, to be multiplied by zero, that's how it's removed. Should I answer the question more clearly. – астон вілла олоф мэллбэрг Sep 10 '16 at 10:32
• Oh I get it thank you very much – user367737 Sep 10 '16 at 10:33
• I wrote an answer anyway. I want to make sure you are satisfied. – астон вілла олоф мэллбэрг Sep 10 '16 at 10:36

$$\cot(X)-\cos(X) = 0 \implies \frac{\cos X}{\sin X} - \cos X = 0 \implies \cos X\bigg(\frac{1}{\sin X} - 1\bigg) = 0$$ $$\implies \frac{\cos X(1 - \sin X)}{\sin X} = 0 \implies \cos X(1 - \sin X) = 0 \cdot \sin X = 0$$
This applies when $\sin X \neq 0$. If it is zero, then $\cot X = \infty$ so the equation is not satisfied.