# If sin square theta-cos square theta=2-5cos theta then what is the value of theta?

Aug 12, 2018

$\theta = 2 k \pi \pm a r c \cos \left(\frac{5 - \sqrt{17}}{4}\right) , k \in \mathbb{Z}$

#### Explanation:

Here ,

${\sin}^{2} \theta - {\cos}^{2} \theta = 2 - 5 \cos \theta , \ldots \ldots .0 \le \theta \le 2 \pi$

$\therefore 1 - {\cos}^{2} \theta - {\cos}^{2} \theta = 2 - 5 \cos \theta$

$\therefore 2 {\cos}^{2} \theta - 5 \cos \theta + 1 = 0$

Comparing with $a {x}^{2} + b x + c = 0$

$a = 2 , b = - 5 \mathmr{and} c = 1$

$\Delta = {b}^{2} - 4 a c = {\left(- 5\right)}^{2} - 4 \left(2\right) \left(1\right) = 17$

So ,

$\cos \theta = \frac{- b \pm \sqrt{\Delta}}{2 a} = \frac{5 \pm \sqrt{17}}{2 \cdot 2}$

$\therefore \cos \theta = \frac{5 - \sqrt{17}}{4} \mathmr{and} \cos \theta = \frac{5 + \sqrt{17}}{4}$

$\therefore \cos \theta \approx 0.22 \mathmr{and} \cos \theta \approx 2.28 \notin \left[- 1 , 1\right]$

$\therefore \cos \theta \approx 0.22 > 0 \implies {1}^{s t} Q u a \mathrm{dr} a n t \mathmr{and} {4}^{t h} Q u a \mathrm{dr} a n t$

:.theta=arc cos((5-sqrt17)/4) or theta=2pi-arc cos((5- sqrt17)/4)

$\therefore \theta = {\left(1.35\right)}^{R}$ OR $\theta = 2 \pi - {\left(1.35\right)}^{R}$ ,where , 0 <= theta < 2pi

The general solution Is :

$\theta = 2 k \pi \pm a r c \cos \left(\frac{5 - \sqrt{17}}{4}\right) , k \in \mathbb{Z}$