MyQuestionIcon

1

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Stationary Point

If sin-1 x+...

Question

If $s i n^{- 1} x + 4 c o s^{- 1} x = π$ , then $x =$

A

1 / 2

No worries! We‘ve got your back. Try BYJU‘S free classes today!

B

\frac{1}{\sqrt{2}}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

C

\frac{\sqrt{3}}{2}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

D

1

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is D $\frac{\sqrt{3}}{2}$
$s i n^{- 1} x + 4 c o s^{- 1} x = π$
$\frac{π}{2} + 3 c o s^{- 1} x = π$
$3 c o s^{- 1} x = \frac{π}{2}$
$c o s^{- 1} x = \frac{π}{6}$
$\Rightarrow x = \frac{\sqrt{3}}{2}$

Suggest Corrections

thumbs-up

0

Similar questions

Q. Inverse circular functions,Principal values of

{s i n}^{- 1} x, {c o s}^{- 1} x, {t a n}^{- 1} x

.

{t a n}^{- 1} x + {t a n}^{- 1} y = {t a n}^{- 1} \frac{x + y}{1 - x y}

x y < 1

π + {t a n}^{- 1} \frac{x + y}{1 - x y}

x y > 1

3 {s i n}^{- 1} \frac{2 x}{1 + x^{2}} - 4 {c o s}^{- 1} \frac{1 - x^{2}}{1 + x^{2}} + 2 {t a n}^{- 1} \frac{2 x}{1 - x^{2}} = \frac{π}{3}

s i n [(1 / 5) {c o s}^{- 1} x] = 1

{t a n}^{- 1} \sqrt{x^{2} + x} + {s i n}^{- 1} \sqrt{x^{2} + x + 1} = \frac{π}{2}

Inverse circular functions,Principal values of

{s i n}^{- 1} x, {c o s}^{- 1} x, {t a n}^{- 1} x

{t a n}^{- 1} x + {t a n}^{- 1} y = {t a n}^{- 1} \frac{x + y}{1 - x y}

x y < 1

π + {t a n}^{- 1} \frac{x + y}{1 - x y}

x y > 1

{c o s}^{- 1} x + {c o s}^{- 1} [\frac{x}{2} + \frac{\sqrt{(3 - 3 x^{2})}}{2}]

(\frac{1}{2} \leq x \leq 1)

c o s (2 {c o s}^{- 1} x + {s i n}^{- 1} x)

x = 1 / 5

0 \leq {c o s}^{- 1} x \leq π

- π / 2 \leq {s i n}^{- 1} x \leq π / 2

4 {cos}^{- 1} x + {sin}^{- 1} x = π

, then the value of

x

{sin}^{- 1} x + {cot}^{- 1} [\frac{1}{2}] = \frac{π}{2}

Inverse circular functions,Principal values of ${sin}^{- 1} x, {c o s}^{- 1} x, {tan}^{- 1} x$ .
${tan}^{- 1} x + {tan}^{- 1} y = {tan}^{- 1} \frac{x + y}{1 - x y}$ , $x y < 1$
$π + {tan}^{- 1} \frac{x + y}{1 - x y}$ , $x y > 1$ .
Evaluate the following :
(a) $sin [\frac{π}{3} - {sin}^{- 1} (- \frac{1}{2})]$
(b) $sin [\frac{π}{2} - {sin}^{- 1} (- \frac{\sqrt{3}}{2})]$

View More

Join BYJU'S Learning Program

similar_icon

Related Videos

lock

Extrema

MATHEMATICS

Watch in App

explore_more_icon

Explore more

Stationary Point

Standard XII Mathematics

Join BYJU'S Learning Program

CrossIcon