Chapter 5, Problem 54CR

To determine

To evaluate: the solution of given trigonometric solution.

Answer to Problem 54CR

The solution of given equation is

  $x=2\pi n,x=\frac{2\pi }{3}+2\pi n,\frac{4\pi }{6}+2\pi n$

Explanation of Solution

Given:

  ${\sin }^{2}x+2\cos x=2$

It is given that

  ${\sin }^{2}x+2\cos x=2$

To rewrite the expression,

  $\begin{array}{l}{\sin }^{2}x+\cos x=2\\ 1-{\cos }^{2}x+\cos x=-1\\ {\cos }^{2}x-\cos x-2=0\end{array}$

Let $\cos x=y$

Further,

  $y^2-y-2=0$

Applying the quadratic equation,

  $\begin{array}{l}y=\frac{1\pm \sqrt{1-4\left(1\right)\left(-2\right)}}{2}\\ y=\frac{1\pm \sqrt{9}}{2}\\ y=\frac{1\pm 3}{4}\\ y=1,-\frac{1}{2}\end{array}$

So,

  $\begin{array}{l}\cos x=1\\ x=2\pi n\end{array}$

Further,

  $\begin{array}{l}\cos x=-\frac{1}{2}\\ x=\frac{2\pi }{3}+2\pi n,\frac{4\pi }{6}+2\pi n\end{array}$

Hence, the solution of given equation is

  $x=2\pi n,x=\frac{2\pi }{3}+2\pi n,\frac{4\pi }{6}+2\pi n$

.