Question

I need some explanations on how to solve these problems. The first two are solving for theta θ.

I could really use help, thanks guys!

Answer 1

6. If you know one solution, you can find all of them.
`\tan\theta=-1` when
`\theta=\frac{3\pi}4`.
`\tan x` has a period of
`\pi`, which means
`\tan(x+\pi)=\tan x`. So we know that
`\theta=\frac{7\pi}4` is another solution. But the same idea applies to this solution, so that
`\theta=\frac{11\pi}4` also works, and so on... The general solution would be
`\theta=\frac{3\pi}4+n\pi` for any integer
`n`.

7. Similar idea. You probably know that
`\sin x=\frac12` when
`x=\frac\pi6`.
`\sin x` has a period of
`2\pi`, so adding any multiple of
`2\pi` to this solution gives more solutions. Here
`x=\frac\theta2`, so we have
`\frac\theta2=\frac\pi6+2n\pi` or
`\theta=\frac\pi3+4n\pi`.

8. This quadratic can be factored:

`2\sin^2x-5\sin x-3=(2\sin x+1)(\sin x-3)=0`

`\implies2\sin x+1=0\text{ or }\sin x=3`

The latter can never happen for real
`x`, since
`-1\le\sin x\le1`.

`2\sin x+1=0\implie\sin x=-\frac12\implies x=-\frac\pi6`

so the general solution would be
`x=-\frac\pi6+2n\pi`.