Solve the equation 2secxsinx+2=4sinx+secx in the interval [0^@,360^@]?

Feb 18, 2017

$x = {129.564}^{\circ}$ and $x = {321.564}^{\circ}$

Explanation:

$2 \sec x \sin x + 2 = 4 \sin x + \sec x$

$\frac{2 \sin x}{\cos x} + 2 = \frac{4 \sin x \cos x + 1}{\cos x}$

$2 \left(\sin x + \cos x\right) = 4 \sin x \cos x + 1$ (1)
- Multiplying both side by cos x (condition cos x diff. to zero)

Call $\left(\sin x + \cos x\right) = u$

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

$= 1 + 2 \sin x \cos x$.

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

$4 \sin x \cos x = 2 {u}^{2} - 2$

Substitute these values into (1):

$2 u = 2 {u}^{2} - 2 + 1$

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

Solve this quadratic equation for $u = \left(\sin x + \cos x\right)$
$D = {d}^{2} = {b}^{2} - 4 a c = 4 + 8 = 12$ --> $d = \pm 2 \sqrt{3}$

There are 2 real roots:
$\sin x + \cos x = u = \frac{2}{2} \pm \frac{2 \sqrt{3}}{4} = 2 \pm \frac{\sqrt{3}}{2}$
${u}_{1} = 1 + \frac{\sqrt{3}}{2}$ (Rejected as $> \sqrt{2}$)
${u}_{2} = 1 - \frac{\sqrt{3}}{2} = 0.134$

$\sin x + \cos x = \sqrt{2} \cos \left(x - \frac{\pi}{4}\right) = 0.134$
$\cos \left(x - \frac{\pi}{4}\right) = \frac{0.134}{\sqrt{2}} = 0.0947$

$x - \frac{\pi}{4} = \pm {84.564}^{\circ}$
$x = 84.564 + 45 = {129.564}^{\circ}$ and
$x = 360 - 84.564 + 45 = {321.564}^{\circ}$