The equation $\sin x\cos x = 2$ has
One solution
Two solutions
Infinite solutions
No solutions
(d) $\sin x\cos x = 2$ or $\sin 2x = 4$, which is impossible.
If $(2\cos x – 1)(3 + 2\cos x) = 0,\,0 \le x \le 2\pi $, then $x = $
$cos (\alpha \,-\,\beta ) = 1$ and $cos (\alpha +\beta ) = 1/e$ , where $\alpha , \beta \in [-\pi , \pi ]$ . Number of pairs of $(\alpha ,\beta )$ which satisfy both the equations is
The number of solution of the given equation $a\sin x + b\cos x = c$ , where $|c|\, > \,\sqrt {{a^2} + {b^2}} ,$ is
If $\sin 2\theta = \cos \theta ,\,\,0 < \theta < \pi $, then the possible values of $\theta $ are
The number of solutions of $tan\, (5\pi\, cos\, \theta ) = cot (5 \pi \,sin\, \theta )$ for $\theta$ in $(0, 2\pi )$ is :
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