# Can you help me with this problem: sinx+cosx=sinxcosx? Please?

Mar 17, 2017

$x = \pi + \frac{1}{2} \arcsin \left(2 - 2 \sqrt{2}\right)$

#### Explanation:

$\sin x + \cos x = \sin x \cos x$

${\left(\sin x + \cos x\right)}^{2} = {\sin}^{2} x {\cos}^{2} x$

${\sin}^{2} x + 2 \sin x \cos x + {\cos}^{2} x = {\sin}^{2} x {\cos}^{2} x$

$1 + 2 \sin x \cos x = {\sin}^{2} x {\cos}^{2} x$

Let $y = \sin x \cos x$

$1 + 2 y = {y}^{2}$

${y}^{2} - 2 y - 1 = 0$

$y = 1 \pm \sqrt{2}$

$\sin x \cos x = 1 \pm \sqrt{2}$

$\sin 2 x = 2 \sin x \cos x \therefore \frac{1}{2} \sin 2 x = \sin x \cos x$

$\frac{1}{2} \sin 2 x = 1 \pm \sqrt{2}$

$\sin 2 x = 2 \pm 2 \sqrt{2}$

$2 + 2 \sqrt{2} > 1 \therefore$ it has no solutions

$\sin 2 x = 2 - 2 \sqrt{2}$

$2 x = \arcsin \left(2 - \sqrt{2}\right)$

$x = \frac{1}{2} \arcsin \left(2 - \sqrt{2}\right)$

Checking this value, we see that the two sides of the equation don't have the same value. This is because $\sin x < 0$ but $\left\mid \cos x \right\mid > \left\mid \sin x \right\mid$, so one side of the eqn is positive whilst the other is negative.

So we add $\pi$ to the value to make $\sin$ positive and $\cos$ negative but we also keep $\left\mid \cos x \right\mid > \left\mid \sin x \right\mid$, so now both sides are negative and, more importantly, have the same value.

So $x = \pi + \frac{1}{2} \arcsin \left(2 - 2 \sqrt{2}\right)$

Apr 17, 2017

$x = \pi + \frac{1}{2} \arcsin \left(2 \left(1 - \sqrt{2}\right)\right)$

#### Explanation:

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

so

${\left(\sin x \cos x\right)}^{2} - 2 \sin x \cos x - 1 = 0$

so

$\sin x \cos x = \frac{2 \pm \sqrt{4 + 4}}{2}$

so

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

but

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

then

$x = \pi + \frac{1}{2} \arcsin \left(2 \left(1 - \sqrt{2}\right)\right)$