Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa

The solution set of the inequality \cos ^2 \theta<\frac{1}{2} is

Option: 1

\begin{aligned} & \left\{\theta:(8 n+1) \frac{\pi}{4}<\theta<(8 n+3) \frac{\pi}{4}, n \in I\right\} \\ \end{aligned}


Option: 2

\begin{aligned} & \left\{\theta:(8 n-3) \frac{\pi}{4}<\theta<(8 n-1) \frac{\pi}{4}, n \in I\right\} \\ \end{aligned}


Option: 3

\begin{aligned} & \left\{\theta:(4 n+1) \frac{\pi}{4}<\theta<(4 n+3) \frac{\pi}{4}, n \in I\right\} \end{aligned}

 


Option: 4

none of the above


Answers (1)

best_answer

\begin{array}{lc} & \cos ^2 \theta<\frac{1}{2} \\ \Rightarrow & 2 \cos ^2 \theta-1<0 \\ \Rightarrow & \cos 2 \theta<0 \\ \Rightarrow & -\cos 2 \theta>0 \\ \Rightarrow & \cos (\pi+2 \theta)>\cos \frac{\pi}{2} \end{array}

\begin{aligned} & \therefore 2 n \pi-\frac{\pi}{2}<\pi+2 \theta<2 n \pi+\frac{\pi}{2}, n \in I \\ & \Rightarrow 2 n \pi-\frac{3 \pi}{2}<2 \theta<2 n \pi-\frac{\pi}{2} \end{aligned}

\begin{array}{cc} \text { or } & n \pi-\frac{3 \pi}{4}<\theta<n \pi-\frac{\pi}{4} \\ \text { or } & (4 n-3) \frac{\pi}{4}<\theta<(4 n-1) \frac{\pi}{4}, n \in I \end{array}

Replacing n by n+1, then
(4 n+1) \frac{\pi}{4}<\theta<(4 n+3) \frac{\pi}{4}, n \in I
 

Posted by

manish painkra

View full answer

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE

Similar Questions

Latest Question