6. If you know one solution, you can find all of them. [tex]\tan\theta=-1[/tex] when [tex]\theta=\dfrac{3\pi}4[/tex]. [tex]\tan x[/tex] has a period of [tex]\pi[/tex], which means [tex]\tan(x+\pi)=\tan x[/tex]. So we know that [tex]\theta=\dfrac{7\pi}4[/tex] is another solution. But the same idea applies to this solution, so that [tex]\theta=\dfrac{11\pi}4[/tex] also works, and so on... The general solution would be [tex]\theta=\dfrac{3\pi}4+n\pi[/tex] for any integer [tex]n[/tex].
7. Similar idea. You probably know that [tex]\sin x=\dfrac12[/tex] when [tex]x=\dfrac\pi6[/tex]. [tex]\sin x[/tex] has a period of [tex]2\pi[/tex], so adding any multiple of [tex]2\pi[/tex] to this solution gives more solutions. Here [tex]x=\dfrac\theta2[/tex], so we have [tex]\dfrac\theta2=\dfrac\pi6+2n\pi[/tex] or [tex]\theta=\dfrac\pi3+4n\pi[/tex].
8. This quadratic can be factored:
[tex]2\sin^2x-5\sin x-3=(2\sin x+1)(\sin x-3)=0[/tex]
[tex]\implies2\sin x+1=0\text{ or }\sin x=3[/tex]
The latter can never happen for real [tex]x[/tex], since [tex]-1\le\sin x\le1[/tex].
[tex]2\sin x+1=0\implie\sin x=-\dfrac12\implies x=-\dfrac\pi6[/tex]
so the general solution would be [tex]x=-\dfrac\pi6+2n\pi[/tex].
