Question

Solve the given equation $7{{\sin }^{2}}\theta +3{{\cos }^{2}}\theta =4$ and find the value of $\theta$

Answer 1

Hint: We will convert ${{\cos }^{2}}\theta$ to ${{\sin }^{2}}\theta$ using the formula ${{\cos }^{2}}x=1-{{\sin }^{2}}x$ and then we will take all the variable to one side and constant to other side. After that we will find the value of $\sin \theta$ and then we will use the formula for the general solution of sin and find the value of $\theta$.

Complete step-by-step answer:
Let’s start solving the question.
We have $7{{\sin }^{2}}\theta +3{{\cos }^{2}}\theta =4$,
Now using the formula ${{\cos }^{2}}x=1-{{\sin }^{2}}x$, to convert cos to sin we get,
$\begin{aligned} & 7{{\sin }^{2}}\theta +3\left( 1-{{\sin }^{2}}\theta \right)=4 \\\ & 7{{\sin }^{2}}\theta +3-3{{\sin }^{2}}\theta =4 \\\ & 4{{\sin }^{2}}\theta =4-3 \\\ & 4{{\sin }^{2}}\theta =1 \\\ & {{\sin }^{2}}\theta =\dfrac{1}{4} \\\ \end{aligned}$
Now we can see that we will get two solution from the above equation,
Therefore, $\sin \theta =\pm \dfrac{1}{2}$
Now first we will find the value of $\theta$ for $\sin \theta =\dfrac{1}{2}$
We know that $\sin \dfrac{\pi }{6}=\dfrac{1}{2}$
Hence, from this we get
$\sin \theta =\sin \dfrac{\pi }{6}$
Now, if we have $\sin \theta =\sin \alpha$ then the general solution is:
$\theta =n\pi +{{\left( -1 \right)}^{n}}\alpha$
Now using the above formula for $\sin \theta =\sin \dfrac{\pi }{6}$ we get,
$\theta =n\pi +{{\left( -1 \right)}^{n}}\dfrac{\pi }{6}$, where n = integers.
Hence, from this we can see that we will get infinitely many solutions for $\theta$.
Now for $\sin \theta =\dfrac{-1}{2}$,
We know that $\sin \dfrac{-\pi }{6}=\dfrac{-1}{2}$.
Now we just have to replace $\dfrac{\pi }{6} by \dfrac{-\pi }{6}$ in $\theta =n\pi +{{\left( -1 \right)}^{n}}\dfrac{\pi }{6}$ and we get,
$\theta =n\pi -{{\left( -1 \right)}^{n}}\dfrac{\pi }{6}$
Hence, solution are $\theta =n\pi +{{\left( -1 \right)}^{n}}\dfrac{\pi }{6}$ and $\theta =n\pi -{{\left( -1 \right)}^{n}}\dfrac{\pi }{6}$.

Note: One can also solve this question by converting sin to cos using the formula ${{\cos }^{2}}x=1-{{\sin }^{2}}x$ and then we will have to use the formula for general solution of cos, and again we will get two solutions and hence, the answer that we get from both the methods are correct and one can use any one of these methods to solve this question.