Free Study Material

NCERT Solutions

NCERT Books

Revision Notes

Practice Problems

This is one of our many quizzes on Solving trigonometric equations. Click on this to access more such competitive quizzes on Trigonometric Functions.

Solving trigonometric equations Quiz 2

mathematics

 Solve the trigonometric equation ${{11}^{\tan \dfrac{v}{2}}}=1.$

 Convert the given equation in the form $\tan x=\tan y$ and then solve

  $\text{Consider the given trigonometric equation }{{11}^{\tan \dfrac{v}{2}}}=1 \\ \text{Write }1\text{ as  }{{11}^{0}}\text{ in R.H.S} \\ \Rightarrow {{11}^{\tan \dfrac{v}{2}}}={{11}^{0}} \\ \tan \dfrac{v}{2}=0\Rightarrow \tan \dfrac{v}{2}=\tan {{0}^{\circ }} \\ \Rightarrow \dfrac{v}{2}=0 \\ \Rightarrow v={{0}^{\circ}}$

 Write the value of which satisfies the trigonometric equation $\sqrt{\dfrac{1+\cos \left( \dfrac{3\theta }{5} \right)}{1-\cos \left( \dfrac{3\theta }{5} \right)}=1}$.

 Use half angle formulas and convert the equation in the form $\cot x=\cot y$ and then solve.

  $\text{The given trigonometric equation is } \\ \sqrt{\dfrac{1+\cos \left( \dfrac{3\theta }{5} \right)}{1-\cos \left( \dfrac{3\theta }{5} \right)}}=1 \\ \Rightarrow \sqrt{\dfrac{2{{\cos }^{2}}\left( \dfrac{3\theta }{10} \right)}{2{{\sin }^{2}}\left( \dfrac{3\theta }{10} \right)}}=1\Rightarrow \cot \left( \dfrac{3\theta }{10} \right)=1 \\ \Rightarrow \cot \left( \dfrac{3\theta }{10} \right)=\cot \dfrac{\pi }{4}\,\,\,\,\,\,\,\,\left[ \because \cot \dfrac{\pi }{4}=1 \right] \\ \Rightarrow \dfrac{3\theta }{10}=\dfrac{\pi }{4} \\ \Rightarrow \theta =\dfrac{5\pi }{6}$

 If $\cos \left( \dfrac{5x}{7} \right)=0$, then which of the following is true?

 Convert the given equation to the form of  $\cos x=\operatorname{cosy}$ and then solve.

  $\text{Given, }\cos \dfrac{5x}{7}=0 \\ \Rightarrow \cos \dfrac{5x}{7}=\cos \left( 2n+1 \right)\dfrac{\pi }{2},n\in Z \\ \dfrac{5x}{7}=\left( 2n+1 \right)\dfrac{\pi }{2},n\in Z \\ \Rightarrow x=\dfrac{7\pi }{10}\left( 2n+1 \right),n\in Z$

 Solve the trigonometric equation $\sin \left( \dfrac{5w+3}{9} \right)=0.$

 First write  $\sin n\pi =0$ for all  $n\in Z$  and then solve.

  $\text{Consider the given equation }\sin \left( \dfrac{5w+3}{9} \right)=0 \\ \text{Write  }0\text{ as }\sin n\pi\text{ for  }n\in Z\text{ is  R.H.S} \\ \Rightarrow \sin \left( \dfrac{5w+3}{9} \right)=\sin n\pi ,n\in Z \\ \Rightarrow \dfrac{5w+3}{9}=n\pi ,n\in Z \\ \Rightarrow 5w+3=9n\pi ,n\in Z\Rightarrow w=\dfrac{9n\pi -3}{5},n\in Z$

 Write the value of $y$ be solving the trigonometric equation ${{\tan }^{-1}}\left( \dfrac{\dfrac{4y}{3}}{1-\dfrac{4{{y}^{2}}}{9}} \right)=2.$

 Reduce the given equation to its simplest form by using suitable trigonometric substitutions. Then solve further.

  $\text{Given, } {{\tan }^{-1}}\left( \dfrac{\dfrac{4y}{3}}{1-\dfrac{4{{y}^{2}}}{9}} \right)=2 \\ \Rightarrow {{\tan }^{-1}}\left( \dfrac{\dfrac{4y}{3}}{1-{{\left( \dfrac{2y}{3} \right)}^{2}}} \right)=2 \\ \text{Put  }\dfrac{2y}{3}=\tan \theta \Rightarrow \theta ={{\tan }^{-1}}\left( \dfrac{2y}{3} \right) \\ \text{The given equation becomes, } \\ {{\tan }^{-1}}\left( \dfrac{2\tan \theta }{1-{{\tan }^{2}}\theta } \right)=2 \\ \Rightarrow {{\tan }^{-1}}\left( \tan 2\theta  \right)=2 \\ \Rightarrow 2\theta =2 \\ \Rightarrow \theta =1 \\ \text{Substitute  }\theta ={{\tan }^{-1}}\dfrac{2y}{3} \\ {{\tan }^{-1}}\dfrac{2y}{3}=1 \\ \Rightarrow {{\tan }^{-1}}\dfrac{2y}{3}=\tan \dfrac{\pi }{4} \\ \Rightarrow \dfrac{2y}{3}=\dfrac{\pi }{4} \Rightarrow y=\dfrac{3\pi }{8}$