Question

How do you solve $2{{\cos }^{2}}x=3\sin x$?

Answer 1

Hint: In this question, we are given an equation in the form of trigonometric function. We need to solve it to find the value of x. For this, we will first change ${{\cos }^{2}}x$ into ${{\sin }^{2}}x$ using the property ${{\sin }^{2}}x+{{\cos }^{2}}x=1$. We will get a quadratic equation in terms of sinx. We will put sinx as a and solve the quadratic equation by using split the middle term method. Using the value of sinx we will then find a particular value siny for which sinx = siny. Then we will use the general solution of x i.e. if sinx = siny then $x=n\pi +{{\left( -1 \right)}^{n}}y$ where $n\in z$.

Complete step-by-step answer:
Here we are given the equation as $2{{\cos }^{2}}x=3\sin x$. We need to find the value of x by solving it.
Let us first change the ${{\cos }^{2}}x$ into ${{\sin }^{2}}x$.
We know that ${{\sin }^{2}}x+{{\cos }^{2}}x=1$ so rearranging it we can say ${{\cos }^{2}}x=1-{{\sin }^{2}}x$. Putting this in our given equation we get, $2\left( 1-{{\sin }^{2}}x \right)=3\sin x\Rightarrow 2-2{{\sin }^{2}}x=3\sin x$.
Rearranging the equation we get $2{{\sin }^{2}}x-3\sin x-2=0$.
We get a quadratic equation in terms of sinx. To solve it easily let us suppose that ${{\sin }^{2}}x=a$. So our equation becomes, $2{{a}^{2}}-3a-2=0$.
Now let us solve this equation to find the value of a = sinx. Comparing $a{{x}^{2}}+bx+c$ we have b = 2, c = -3, d = -2. We want ${{n}_{1}},{{n}_{2}}$ such that ${{n}_{1}}+{{n}_{2}}=-3\text{ and }{{n}_{1}}\cdot {{n}_{2}}=b\cdot d=-4$.
We have ${{n}_{1}}=-4,{{n}_{2}}=1$.
So splitting the middle term we have \[2{{a}^{2}}-4a+a-2=0\].
Taking 2a common from first two terms and 1 from last two terms we get $2a\left( a-2 \right)+\left( a-2 \right)=0\Rightarrow \left( 2a+1 \right)\left( a-2 \right)=0$.
This implies that 2a+1 = 0 and a-2 = 0, $a=\dfrac{-1}{2}\text{ and }a=2$.
Putting a sinx we get $\sin x=\dfrac{-1}{2}\text{ and }\sin x=2$.
We know that, value of sinx lies between -1 and 1 only, so sinx cannot be equal to 2.
Therefore, we have $\sin x=\dfrac{-1}{2}$.
We know that $\sin \dfrac{\pi }{6}=\dfrac{1}{2}$.
But we need a negative of $\dfrac{1}{2}$. So we need to look at the quadrant, we know sine is negative in the third quadrant so we can say $\sin \left( \pi +\theta \right)=-\sin \theta $.
Taking $\theta $ as $\dfrac{\pi }{6}$ we have $\sin \left( \pi +\dfrac{\pi }{6} \right)=-\dfrac{1}{2}\Rightarrow \sin \dfrac{7\pi }{6}=-\dfrac{1}{2}$.
So one of the values of x will be $\dfrac{7\pi }{6}$. But let us find the general solution. We know that, when sinx = siny $\Rightarrow x=n\pi +{{\left( -1 \right)}^{n}}y$.
Here we have $\sin x=\sin \dfrac{7\pi }{6}$ where $n\in z$.
So $x=n\pi +{{\left( -1 \right)}^{n}}\dfrac{7\pi }{6}$ where $n\in z$.
Therefore, our required values of x are $n\pi +{{\left( -1 \right)}^{n}}\dfrac{7\pi }{6}$ where n belong to the set of integers.

Note: Students often make mistakes of finding just one value of x which will give partial marks only. We need to find all values of x which satisfy the given equation. Take care of the signs while solving the quadratic equation by splitting the middle term method.