# Solve the equation cos^2x-sin^2x=sinx+1 in the interval (0<=x<=2pi)?

Oct 4, 2017

$x = \left\{0 , \frac{5 \pi}{6} , \pi , \frac{7 \pi}{6} , 2 \pi\right\}$

#### Explanation:

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

we have $1 - {\sin}^{2} x - {\sin}^{2} x = \sin x + 1$

or $2 {\sin}^{2} x + \sin x = 0$

or $\sin x \left(2 \sin x + 1\right) = 0$

i.e. either $\sin x = 0 = \sin 0$ i.e. $x = n \pi$

or $\sin x = - \frac{1}{2} = \sin \left(- \frac{\pi}{6}\right)$ i.e. $x = n \pi + {\left(- 1\right)}^{n} \left(- \frac{\pi}{6}\right)$

and with in period $\left[0 , 2 \pi\right]$, we have

$x = \left\{0 , \frac{5 \pi}{6} , \pi , \frac{7 \pi}{6} , 2 \pi\right\}$

Oct 4, 2017

$x = 0 , \frac{5 \pi}{6} , \pi , \frac{7 \pi}{6} , 2 \pi$ $\left(0 \le x \le 2 \pi\right)$

#### Explanation:

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

Here we the above equation to get two new equations (I'm assuming you know how to solve quadratic equations):

$\sin x = 0$
$\therefore x = 0 , \pi , 2 \pi$ $\left(0 \le x \le 2 \pi\right)$

OR

$2 \sin x + 1 = 0$
$\sin x = - \frac{1}{2}$

Here, we cannot go straight to saying $x = \frac{\pi}{6}$, as this would give a positive answer. Sine is negative in the 3rd and 4th quadrants, so these of $x$ in $\sin x = - \frac{1}{2}$ can be found by putting the acute positive angle $x = \frac{\pi}{6}$ into $\pi + x$, and $2 \pi - x$.

So,

$2 \sin x + 1 = 0$
$\sin x = - \frac{1}{2}$
$x = \pi + \frac{\pi}{6} , 2 \pi - \frac{\pi}{6}$
$\therefore x = \frac{5 \pi}{6} , \frac{7 \pi}{6}$ $\left(0 \le x \le 2 \pi\right)$

Putting these two sets of answers together:
$\therefore x = 0 , \frac{5 \pi}{6} , \pi , \frac{7 \pi}{6} , 2 \pi$ $\left(0 \le x \le 2 \pi\right)$

However this is assuming that the domain the question asked for is inclusive of $0$ and $2 \pi$.

If it is exclusive of these, the solutions to this equation will be:
$x = \frac{5 \pi}{6} , \pi , \frac{7 \pi}{6}$ $\left(0 < x < 2 \pi\right)$

Solving this is just like solving an algebraic equation once you get everything in terms of $\sin x$, so be sure to know your algebra and trig identities back to front!