MCQ
If $4{\sin ^{ - 1}}x + {\cos ^{ - 1}}x = \pi ,$ then $x$ is equal to
  • A
    $0$
  • $\frac{1}{2}$
  • C
    $ - \frac{{\sqrt 3 }}{2}$
  • D
    $\frac{1}{{\sqrt 2 }}$

Answer

Correct option: B.
$\frac{1}{2}$
b
(b) We know that $4{\sin ^{ - 1}}x + {\cos ^{ - 1}}x = \pi $

==> $3{\sin ^{ - 1}}x + {\sin ^{ - 1}}x + {\cos ^{ - 1}}x = \pi $

==> $3{\sin ^{ - 1}}x = \pi - \frac{\pi }{2} = \frac{\pi }{2}$

==> ${\sin ^{ - 1}}x = \pi /6$

==> $x = \sin \frac{\pi }{6} = \frac{1}{2}$.

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free

Similar questions

If $y = y ( x ), x \in\left(0, \frac{\pi}{2}\right)$ be the solution curve of the differential equation $\left(\sin ^{2} 2 x\right) \frac{d y}{d x}+\left(8 \sin ^{2} 2 x+2 \sin 4 x\right) y=$$2 e ^{-4 x }(2 \sin 2 x +\cos 2 x )$, with $y \left(\frac{\pi}{4}\right)= e ^{-\pi}$, then $y \left(\frac{\pi}{6}\right)$ is equal to.
if $\text{y}=\text{e}^{{\tan}\text{x}},$ then $(\cos^2\text{x})\text{y}_2=$
Let $\vec a $ and $\vec b $ be two unit vectors. If the vector $\vec c=\vec a+2\vec b $ and $\vec d=5\vec a-4\vec b $ are perpendicular to each other, then the angle between $\vec a$  and $\vec b$  
The area enclosed by the curve $y={{\sin }^{-1}}\left( \cos x \right)$ and $y={{\cos }^{-1}}\left( \sin x \right)$ for $x\in \left[ \frac{\pi }{2},\frac{3\pi }{2} \right]$, is
The area bounded by $y –1 = |x|, y = 0$ and $|x| \frac{1}{2}$ will be:
Every invertible function is:
What is the degree of the differential equation $y =x \frac{d y}{d x}+\left(\frac{d y}{d x}\right)^{-1} ?$
If $\alpha \neq \mathrm{a}, \beta \neq \mathrm{b}, \gamma \neq \mathrm{c}$ and $\left|\begin{array}{lll}\alpha & \mathrm{b} & \mathrm{c} \\ \mathrm{a} & \beta & \mathrm{c} \\ \mathrm{a} & \mathrm{b} & \gamma\end{array}\right|=0$,then $\frac{a}{\alpha-a}+\frac{b}{\beta-b}+\frac{\gamma}{\gamma-c}$ is equal to :
If $\left[ {\begin{array}{*{20}{c}}1&{\,\,1}&{\,\,1}\\1&{ - 2}&{ - 2}\\1&{\,\,3}&{\,\,1}\end{array}} \right]\,\left[ \begin{array}{l}x\\y\\z\end{array} \right] = \left[ \begin{array}{l}0\\3\\4\end{array} \right]$, then $\left[ \begin{array}{l}x\\y\\z\end{array} \right]$is equal to
A vector which bisects the angle between $a =3 i -4 k$ and $b =5 j +12 k$ is