Mind the Interval!

Firstly, we square both the sides to get $1+\sin2x=\dfrac{1}{4}$

$\implies \sin2x=\dfrac{-3}{4}$

Now, we use the identity $\sin2x=\dfrac{2tanx}{1+\tan^{2}x}$ to get a quadratic equation

$3\tan^{2}x+8tanx+3=0$

$\implies tanx=\dfrac{-4±\sqrt{7}}{3}$

Now, we prove that only one of these solutions works.

First of all, to get a better picture, we take approximations.

$\sqrt{7} \approx 2.7$ and thus $\dfrac{-4+\sqrt{7}}{3} \approx \dfrac{-1.3}{3}$ and similarly

$\dfrac{-4-\sqrt{7}}{3} \approx \dfrac{-6.7}{3}$

Now, we know that $tanx<0 \implies \dfrac{π}{2}<x<π$ (since $x \in [0,π])$ $\implies sinx>0$ and $cosx<0$

But $sinx + cosx =\dfrac{1}{2} >0$ .

Thus,

$sinx>|cosx|$

This means that the numerator has to be greater than the denominator in $|tanx|$ and thus

$tanx=\boxed {\dfrac{-(4+\sqrt{7})}{3}}$

I wasnt sure as to how to draw a graph on the comp., so i had to go with images.

By squaring $\sin x + \cos x = \frac{1}{2}$ one has $\sin x \cos x = - \frac{3}{8}$ .

So, $\sin x$ and $\cos x$ are the roots of $8y^2-4y-3=0$ .

As $0 < x < \pi$ , we have $\sin x = \frac{1+\sqrt{7}}{4}$ and $\cos x = \frac{1-\sqrt{7}}{4}$ .

Finally, $\tan x = -\frac{4+\sqrt{7}}{3}$ .

since \sinx+\cosx=\sqrt{2}sin[x+\frac{\pi}{4}]=\frac{1]{2}, we can think of the triangle below. {: .center} \(AE=2\sqrt{2}, AB=1, BE=sqrt{7} and thus $\tan\theta=frac{1}{sqrt{7}}$ .

We have to find \tanx=-\tan(\theta+\frac{\pi}{4}=\frac(1+\tan\theta}{1-\tan\theta|=-\frac{4+\sqrt{7}}{3}