Question

How do you solve $2{{\cos }^{2}}x+3\sin x=3$ in the interval $0\le x\le 2\pi $ ?

Answer 1

Hint: We have been given an inequation consisting of two trigonometric functions, sine of x and cosine of x. Since the mentioned interval is $0\le x\le 2\pi $, therefore, we shall only find the principal values only for which this equation holds true. Firstly, we shall convert the entire equation in terms of the cosine function using the trigonometric identities. Then we shall solve it further like a quadratic function which would be in terms of sinx.

Complete step by step solution:
Given that $2{{\cos }^{2}}x+3\sin x=3$
We shall use the trigonometric identity ${{\sin }^{2}}x+{{\cos }^{2}}x=1$ to substitute the square of cosine of x with 1 minus the square of sine of x, that is, ${{\cos }^{2}}x=1-{{\sin }^{2}}x$.
$\Rightarrow 2\left( 1-{{\sin }^{2}}x \right)+3\sin x=3$
$\Rightarrow 2-2{{\sin }^{2}}x+3\sin x=3$
Transposing the constant term 3 to the left hand side of equation, we get
$\Rightarrow 2-2{{\sin }^{2}}x+3\sin x-3=0$
$\Rightarrow -2{{\sin }^{2}}x+3\sin x-1=0$
We will multiply the entire equation with a negative sign.
$\Rightarrow 2{{\sin }^{2}}x-3\sin x+1=0$
Now, we shall factor the quadratic equation formed above and further group them into two groups.
$\Rightarrow 2{{\sin }^{2}}x-2\sin x-\sin x+1=0$
$\Rightarrow 2\sin x\left( \sin x-1 \right)-1\left( \sin x-1 \right)=0$
$\Rightarrow \left( \sin x-1 \right)\left( 2\sin x-1 \right)=0$
Thus, $\sin x-1$and $2\sin x-1=0$.
For $\sin x-1$,
$\Rightarrow \sin x=1$
This holds true for $x=\dfrac{\pi }{2}$ in the interval $0\le x\le 2\pi $.
For $2\sin x-1=0$,
$\Rightarrow \sin x=\dfrac{1}{2}$
This holds true for $x=\dfrac{\pi }{6},\dfrac{5\pi }{6}$ in the interval $0\le x\le 2\pi $.
Therefore, the solution of $2{{\cos }^{2}}x+3\sin x=3$ in the interval $0\le x\le 2\pi $ is $x=\dfrac{\pi }{6},\dfrac{\pi }{2},\dfrac{5\pi }{6}$.

Note: In order to find the solution of various trigonometric equations, we must have prior knowledge of the main trigonometric identities. Also, we could have substituted the quadratic equation formed in terms of sine of x function with any variable-u to proceed with solving the quadratic equation. Later we could equate the calculated values of u-variable equal to sin x to find our final solution.