Question

How do you solve $2{\cos ^2}x + \sin x - 2 = 0$ over the interval 0 to 2$\pi$?

Answer 1

Hint: First of all, we will convert ${\cos ^2}x$ in terms of ${\sin ^2}x$, then we will get a quadratic equation in $\sin x$ and we will then assume $y = \sin x$ and thus we get the value of $y$.

Complete step by step solution:
We are given that we are required to solve $2{\cos ^2}x + \sin x - 2 = 0$ over the interval 0 to 2$\pi$.
Since, we know that for any real number $\theta $, we have: ${\sin ^2}\theta + {\cos ^2}\theta = 1$.
Replacing $\theta $ by $x$, we will then obtain the following equation with us:-
$ \Rightarrow {\sin ^2}x + {\cos ^2}x = 1$
Taking ${\sin ^2}x$ from addition in the left hand side to subtraction in the right hand side, we will then obtain the following equation with us:-
$ \Rightarrow {\cos ^2}x = 1 - {\sin ^2}x$
Putting this value in the given equation, we will then obtain the following equation with us:-
$ \Rightarrow 2\left( {1 - {{\sin }^2}x} \right) + \sin x - 2 = 0$
Simplifying the brackets on the left hand side of the above equation, we will then obtain the following equation with us:-
$ \Rightarrow 2 - 2{\sin ^2}x + \sin x - 2 = 0$
Simplifying the left hand side of the above equation by combining the constants, we will then obtain the following equation:-
$ \Rightarrow - 2{\sin ^2}x + \sin x = 0$
Multiplying the above equation by – 1, we will then obtain the following equation with us:-
$ \Rightarrow 2{\sin ^2}x - \sin x = 0$
Taking $\sin x$ common from the left hand side in the above equation, we will then obtain the following equation with us:-
$ \Rightarrow 2\sin x\left( {\sin x - 1} \right) = 0$
$ \Rightarrow $ Either $2\sin x = 0$ or $\sin x - 1 = 0$
$ \Rightarrow $ Either $\sin x = 0$ or $\sin x = 1$
$ \Rightarrow $ Either $x = 0,\pi $ or $x = \dfrac{\pi }{2}$

Thus, the possible values of $x$ are $0,\pi ,\dfrac{\pi }{2}$.

Note: The students must note that the underlying fact used to solve $2\sin x\left( {\sin x - 1} \right) = 0$ is as follows:
If a.b = 0, then either a = 0 or b = 0 or both.
The students must commit to memory the following fact: For any real number $\theta $, we have: ${\sin ^2}\theta + {\cos ^2}\theta = 1$.
It can also be proved using the fact that sine of any angle is given by the ratio of perpendicular and the hypotenuse and cosine of any angle is given by the ratio of base and hypotenuse. And, finally we will use the Pythagorean Theorem to prove it equal to 1.