First slide

Trigonometric equations

Question

The equation $(\cos p - 1) x^{2} + (\cos p) x + \sin p = 0$ in the variable x has real roots. Then p can take any value in the interval

Easy

A

$(0, 2 π)$
B

$(- π, 0)$
C

$(\frac{- π}{2}, \frac{π}{2})$
D

$[0, π]$

Solution

$S i n c e t h e e q u a t i o n (\cos p - 1) x^{2} + (\cos p) x + \sin p = 0 h a s r e a l r o o t s \Rightarrow b^{2} - 4 a c \geq 0$
$\Rightarrow \cos^{2} p - 4 \sin p (\cos p - 1) \geq 0$
$\Rightarrow \sin c e \cos^{2} p \geq 0 a n d \cos p - 1 \leq 0$ ,
One of the possibilities for the inequality to hold good is $\sin p \geq 0$
$\Rightarrow p \in [0, π]$

Get Instant Solutions

When in doubt download our app. Now available Google Play Store- Doubts App

Recieve an sms with download link

Doubts App

OR

SCAN ME

Doubts App

Doubts App