Question

Solve the given trigonometric equation for $\theta$:-
$\sqrt{3}\sec 2\theta =2$.

Answer 1

- Hint: We can simplify the above equation as follows

& \Rightarrow \sqrt{3}\sec 2\theta =2 \\\ & \Rightarrow \sec 2\theta =\dfrac{2}{\sqrt{3}} \\\ \end{aligned}$$ In this question, we can use the basic relations that exist among the different trigonometric function which are mentioned as follows $$\begin{aligned} & \Rightarrow {{\sec }^{2}}x-{{\tan }^{2}}x=1 \\\ & \Rightarrow \sec x=\dfrac{1}{\cos x} \\\ \end{aligned}$$ _Complete step-by-step solution_ - Now, we can use the equation which contains the $$\sec 2\theta $$ function that is given above in the question to solve it and get the values/value of $$\theta $$. Also, the relation between $$\cos 2\theta \ and\ \cos \theta $$ is very important to solve this question which is as follows $$\Rightarrow \cos 2\theta =2{{\cos }^{2}}\theta -1$$ One more important thing is that for solving a trigonometric equation, we need to write as follows Let us say cos x = y Then, for solving this equation, we will write as follows $$\begin{aligned} & x=2n\pi \pm {{\cos }^{-1}}y \\\ & \left( n\in I\ and\ y\in \left[ 0,\pi \right] \right) \\\ \end{aligned}$$ As mentioned in the question, we have to find the value/values of $$\theta $$ . Now, we can use any of the two relations for converting sec function into either cos function or tan function and hence, we can write the following On using the cos and sec functions relation, we can write as following $$\begin{aligned} & \Rightarrow \sec x=\dfrac{1}{\cos x} \\\ & \Rightarrow \sec 2\theta =\dfrac{1}{\cos 2\theta } \\\ & \Rightarrow \dfrac{2}{\sqrt{3}}=\dfrac{1}{\cos 2\theta } \\\ & \Rightarrow \cos 2\theta =\dfrac{\sqrt{3}}{2} \\\ \end{aligned}$$ Now, we will use the relation between $$\cos 2\theta \ and\ \cos \theta $$ and hence, we can write as follows $$\begin{aligned} & \Rightarrow \cos 2\theta =2{{\cos }^{2}}\theta -1 \\\ & \Rightarrow \dfrac{\sqrt{3}}{2}=2{{\cos }^{2}}\theta -1 \\\ & \Rightarrow 2{{\cos }^{2}}\theta =1+\dfrac{\sqrt{3}}{2} \\\ & \Rightarrow {{\cos }^{2}}\theta =\dfrac{2+\sqrt{3}}{4} \\\ \end{aligned}$$ Now, on taking the square root on both the sides, we will get the following $$\begin{aligned} & \Rightarrow \cos \theta =\sqrt{\dfrac{2+\sqrt{3}}{4}} \\\ & \Rightarrow \cos \theta =\dfrac{\sqrt{2+\sqrt{3}}}{2} \\\ \end{aligned}$$ Now, for solving this above equation, we can write as follows $$\begin{aligned} & \theta =2n\pi \pm {{\cos }^{-1}}\left( \dfrac{\sqrt{2+\sqrt{3}}}{2} \right) \\\ & \left( n\in I\ and\ \left( \dfrac{\sqrt{2+\sqrt{3}}}{2} \right)\in \left[ 0,\pi \right] \right) \\\ \end{aligned}$$ Hence, this is the solution of the equation. Note:- Another method of doing this question is that- We could have taken the tan and sec functions relation as mentioned in the hint. From the relation, we can get the value of $$\tan 2\theta $$ and then we could have found out the solution similar to the way we did for cos function. Also, the students can commit calculation errors, so, one should be extra careful while solving the question and while doing the calculations.