The number of solution of the given equation | vedclass

Name: PDF Generator App | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(d) $\frac{a}{{\sqrt {{a^2} + {b^2}} }}\sin x + \frac{b}{{\sqrt {{a^2} + {b^2}} }}\cos x = \frac{c}{{\sqrt {{a^2} + {b^2}} }}$

$\sin (x + \alpha )

Vedclass · Accepted Answer

None of these

Vedclass · Answer

(d) $\frac{a}{{\sqrt {{a^2} + {b^2}} }}\sin x + \frac{b}{{\sqrt {{a^2} + {b^2}} }}\cos x = \frac{c}{{\sqrt {{a^2} + {b^2}} }}$

$\sin (x + \alpha ) = \frac{c}{{\sqrt {{a^2} + {b^2}} }} > 1$, (as given)

Hence there is no solution.

The number of solution of the given equation $a\sin x + b\cos x = c$ , where $|c|\, > \,\sqrt {{a^2} + {b^2}} ,$ is

Similar Questions

If ${\tan ^2}\theta - (1 + \sqrt 3 )\tan \theta + \sqrt 3 = 0$, then the general value of $\theta $ is

The number of values of $\theta $ in $[0, 2\pi]$ satisfying the equation $2{\sin ^2}\theta = 4 + 3$$\cos \theta $ are

$\cot \theta = \sin 2\theta (\theta \ne n\pi $, $n$ is integer), if $\theta = $

The number of pairs $(x, y)$ satisfying the equations $\sin x + \sin y = \sin (x + y)$ and $|x| + |y| = 1$ is

Find the general solution of the equation $\sec ^{2} 2 x=1-\tan 2 x$