Why is $-\sqrt{\left(2 \right)} ,\sqrt{\left(2 \right)}$ the interval of the following equation?

Question

Alright,

$(3+\sqrt2)(sinx+cosx) - 2 sinxcosx = 3 \sqrt2 +1 $ is the exercise.

I have the solution of this problem which is the following:

$\sin x+ \cos x=t $
$\sin (x) \cdot \cos (x)=\frac{t^2-1}{6}$

$t\in(-\sqrt2,\sqrt2)$

Can someone explain me why is t limited with that interval?

Thanks.

Answer 1

See that

$$\sin x + \cos x = \sqrt{2}\left(\frac{1}{\sqrt{2}}\sin x +\frac{1}{\sqrt{2}}\cos x \right)$$

$$=\sqrt{2}(\cos \pi/4 \sin x + \sin \pi/4 \cos x )=\sqrt{2}\sin(x+\pi/4).$$

Since $\sin (x+ \pi/4)$ is between $-1$ and $1$, the original expression is between $-\sqrt{2}$ and $\sqrt{2}$ and this is your value for $t$.