Question

How do I solve sinxcotx-cos^2x=0 and find the principal values and all values??

Answer 1

As long as
`\sin x\\eq0`, we have

`\sin x\cot x-\cos^2x=\sin x(\cos x)/(\sin x)-\cos^2x=\cos x-\cos^2x=0`

From here, you can pull out one factor of
`\cos x`:

`\cos x(1-\cos x)=0`

Then either
`\cos x=0` or
`\cos x=1`.

In the first case,
`\cos x=0` whenever
`x=\pm\frac\pi2,\pm\frac{3\pi}2,\pm\frac{5\pi}2,\ldots`, or every odd multiple of
`\frac\pi2`. We denote this by
`x=\frac{(2n+1)\pi}2` for
`n\in\mathbb Z` (
`n` is any integer).

In the second case,
`\cos x=1` whenever
`x=0,\pm2\pi,\pm4\pi,\ldots`, or every even multiple of
`\pi`. We write this as
`x=2n\pi` for
`n\in\mathbb Z`.

I'm not sure what you mean by "principal values", but I'd guess it refers to any solutions to the equation for
`0\le x<2\pi`. In that case, you'd have only 3 solutions,
`x=\frac\pi2,\frac{3\pi}2,0`.

However, we have to throw out the solution
`x=0`, because that makes the left hand side of the original equation undefined. So the general solution is actually
`x=\frac{(2n+1)\pi}2` for
`n\in\mathbb Z` and
`x=2n\pi` for
`n\in\mathbb Z\setminus\{0\}` (all non-zero integers).