Solve $$\tan(x) = 1 - 2\sin(x)$$ for the domain of $[0, \pi/4].$
Thanks! And please just point me to some direction without the answer.
I actually had tried few trig identities and managed to rewrite this as a 4th degree polynomial, but then I did not know how to solve the 4th degree polynomial so I just used my calculator to find the roots of the polynomial, and plugged the roots into arcsin.
Consider that you look for the zero(s) of function $$f(x)=\tan(x) + 2\sin(x)-1$$ By inspection, $f(0)=-1$ and $f\left(\frac{\pi }{4}\right)=\sqrt{2}$. On the other side, the derivative is positive in the whole range. So, the solution is unique.
Assuming that you do not know the value of the trigonometric functions of $\frac{\pi }{8}$, we cannot bound more. But $\frac{\pi }{4}$ is quite small; so, may be, a very truncated Taylor series will provide a very simple approximation.
If you want to polish it, use Newton method.