Question

Solve the following equation $5{{\cos }^{2}}\theta +7{{\sin }^{2}}\theta -6=0$.

Answer 1

Hint: In the above question we will use the trigonometric identity ${{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1$ and then we will simplify it to get the solution of the equation.

Complete Step-by-step answer:
We have been given the equation $5{{\cos }^{2}}\theta +7{{\sin }^{2}}\theta -6=0$.
As we know that ${{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1$.
Also, we can say that ${{\cos }^{2}}\theta =1-{{\sin }^{2}}\theta$.
So, by substituting the value of ${{\cos }^{2}}\theta$ in the above equation, we get as follows:

& 5\left( 1-{{\sin }^{2}}\theta \right)+7{{\sin }^{2}}\theta -6=0 \\\ & 5-5{{\sin }^{2}}\theta +7{{\sin }^{2}}\theta -6=0 \\\ \end{aligned}$$ On adding the similar term in the equation, we get as follows: $$\begin{aligned} & 7{{\sin }^{2}}\theta -5{{\sin }^{2}}\theta +5-6=0 \\\ & 2{{\sin }^{2}}\theta -1=0 \\\ \end{aligned}$$ On adding 1 to both the sides of the equality, we get as follows: $$\begin{aligned} & 2{{\sin }^{2}}\theta -1+1=1 \\\ & 2{{\sin }^{2}}\theta =1 \\\ \end{aligned}$$ On dividing the equation by 2 to both sides of equality, we get as follows: $$\begin{aligned} & \dfrac{2{{\sin }^{2}}\theta }{2}=\dfrac{1}{2} \\\ & {{\sin }^{2}}\theta =\dfrac{1}{2} \\\ \end{aligned}$$ As we know the value of $$\sin \dfrac{\pi }{4}=\dfrac{1}{\sqrt{2}}\Rightarrow {{\sin }^{2}}\dfrac{\pi }{4}=\dfrac{1}{2}$$. So after using the value of $${{\sin }^{2}}\dfrac{\pi }{4}$$ in the above equation, we get as follows: $${{\sin }^{2}}\theta ={{\sin }^{2}}\dfrac{\pi }{4}$$ As we know that the general solution for $${{\sin }^{2}}\theta ={{\sin }^{2}}\alpha $$ is given by $$\theta =n\pi \pm \alpha $$ where ‘n’ is any integer. $$\begin{aligned} & {{\sin }^{2}}\theta ={{\sin }^{2}}\dfrac{\pi }{4} \\\ & \theta =n\pi \pm \dfrac{\pi }{4} \\\ \end{aligned}$$ On substituting the values of the obtained solution in the given equation we can observe that it satisfies all the values of ‘n’. Therefore, the solution for the above question is given by $$\theta =n\pi \pm \dfrac{\pi }{4}$$ where n is any integer. Note: Be careful while substituting the value of $${{\cos }^{2}}\theta $$ in terms of $$si{{n}^{2}}\theta $$ as there is a chance that you put the wrong sign. Also, check the solution once by substituting the values in the equation. We can also take the square root to both sides of the equation $$si{{n}^{2}}\theta =\dfrac{1}{2}$$ but we will have to generalize the solution for $$sin\theta =\dfrac{1}{\sqrt{2}}$$ and $$sin\theta =-\dfrac{1}{\sqrt{2}}$$.