Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa

The equation a\cos x-3\sin x = a+1 is solvable only if a belongs to the interval

  • Option 1)

    \left [a, +\infty \right ]

  • Option 2)

    \left [ -4, 4 \right ]

  • Option 3)

    (-\infty, 4)

  • Option 4)

    None of these

 

Answers (1)

best_answer

As learnt in

Results from Submultiples of an angle -

Results from submultiples of an angle 2

-

 

 a\cos x -3 \sin x = a+1

a( \frac{1- \tan ^{2}\frac{n}{2}}{1+\tan ^{2}\frac{n}{2}}) - 3. \frac{2 \tan \frac{x}{2}}{1+ \tan ^{2} \frac{x}{2}}= a+1

\omega \tan \frac{x}{2}=y

a \left ( \frac{1-y^{2}}{1+y^{2}} \right )- \frac{6y}{1+y^{2}}=a+1

\therefore a -ay^{2}-6y=\left ( 1+y^{2} \right )\left ( a+1 \right )

\therefore y^{2} (2a+1)+6y+1=0

for real y, D\geqslant O

\therefore 36-4 \left ( 2a+1 \right )\geqslant 0

\because a \epsilon (-\infty, +4)

 


Option 1)

\left [a, +\infty \right ]

Incorrect option

Option 2)

\left [ -4, 4 \right ]

Incorrect option

Option 3)

(-\infty, 4)

Correct option

Option 4)

None of these

Incorrect option

Posted by

divya.saini

View full answer

Similar Questions

Latest Question