Let the inverse trigonometric functions take principal values. The number of real solutions of the equation is

Question

Let the inverse trigonometric functions take principal values. The number of real solutions of the equation $2 \sin^{- 1} x + 3 \cos^{- 1} x = \frac{2 π}{5},$ is _____. [2024]

Smart Achievers · Accepted Answer

(0)

We have, $2 \sin^{- 1} x + 3 \cos^{- 1} x = \frac{2 π}{5}$

$\Rightarrow 2 (\frac{π}{2} - \cos^{- 1} x) + 3 \cos^{- 1} x = \frac{2 π}{5}$

$\Rightarrow π + \cos^{- 1} x = \frac{2 π}{5} \Rightarrow \cos^{- 1} x = \frac{- 3 π}{5}$

Which is not possible as $\cos^{- 1} x \in [0, π]$

$∴ No solution exists.$

Solution

Q.
Let the inverse trigonometric functions take principal values. The number of real solutions of the equation $2 \sin^{- 1} x + 3 \cos^{- 1} x = \frac{2 π}{5},$ is _____. [2024]

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Solution

Q. Let the inverse trigonometric functions take principal values. The number of real solutions of the equation 2sin-1x+3cos-1x=2π5, is _____. [2024]

Ans. (0) We have, 2sin-1x+3cos-1x=2π5 ⇒2(π2-cos-1x)+3cos-1x=2π5 ⇒π+cos-1x=2π5 ⇒ cos-1x=-3π5 Which is not possible as cos-1x∈[0,π] ∴No solution exists.

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Q.
Let the inverse trigonometric functions take principal values. The number of real solutions of the equation $2 \sin^{- 1} x + 3 \cos^{- 1} x = \frac{2 π}{5},$ is _____. [2024]