Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • Engineering
  • The number of solution (s) of the equation<img alt="\sin ^3 x \cos x+\sin ^2 x \cos ^2 x+\sin x \cos ^3 x=1" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Csin%20%5E3%20x%20%5Cco

The number of solution (s) of the equation\sin ^3 x \cos x+\sin ^2 x \cos ^2 x+\sin x \cos ^3 x=1 in the interval [0,2 \pi] is/are

Option: 1

 no
 


Option: 2

one
 


Option: 3

two
 

 


Option: 4

three


Answers (1)

best_answer

We have

\begin{aligned} & \sin ^3 x \cos x+\sin ^2 x \cos ^2 x+\sin x \cos ^3 x=1 \\ \Rightarrow & \sin x \cos x\left(\sin ^2 x+\sin x \cos x+\cos ^2 x\right)=1 \end{aligned}

\Rightarrow \quad \frac{\sin 2 x}{2}\left(1+\frac{\sin 2 x}{2}\right)=1

\quad \begin{gathered} \sin 2 x(2+\sin 2 x)=4 \\ \Rightarrow \quad \sin ^2 2 x+2 \sin 2 x-4=0 \end{gathered}

                                            \sin 2 x=\frac{-2 \pm \sqrt{(4+16)}}{2}
\therefore \quad \sin 2 x=-1 \pm \sqrt{5}

Impossible (\because-1 \leq \sin \cdot 2 x \leq 1)

Hence, given equation has no solutions

 

Posted by

Suraj Bhandari

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