Let the inverse trigonometric functions take principal values. The number of real solutions of the equation \[ 2 \sin^{-1} x + 3 \cos^{-1} x = \frac{2\pi}{5}, \] is ______ .
Correct Answer: 0
Approach Solution - 1
To solve the equation \(2 \sin^{-1} x + 3 \cos^{-1} x = \frac{2\pi}{5}\), we start by noting a key identity: \(\sin^{-1} x + \cos^{-1} x = \frac{\pi}{2}\). Let's assign variables for simplicity:
- Let \(a = \sin^{-1} x\) and \(b = \cos^{-1} x\).
According to the identity, \(a + b = \frac{\pi}{2}\).
The given equation is:
\[2a + 3b = \frac{2\pi}{5}\]
Substitute \(b = \frac{\pi}{2} - a\) from the identity:
\[2a + 3\left(\frac{\pi}{2} - a\right) = \frac{2\pi}{5}\]
Expanding this, we get:
\[2a + \frac{3\pi}{2} - 3a = \frac{2\pi}{5}\]
\[-a + \frac{3\pi}{2} = \frac{2\pi}{5}\]
Rearranging gives:
\[-a = \frac{2\pi}{5} - \frac{3\pi}{2}\]
Simplify to find:
\[-a = \frac{2\pi - 7.5\pi}{5} = -\frac{5.5\pi}{5}\]
Thus:
\[a = \frac{5.5\pi}{5} = \frac{11\pi}{10}\]
Now, since \(a = \sin^{-1} x\), it must be true that \(-\frac{\pi}{2} \leq a \leq \frac{\pi}{2}\). However, \(\frac{11\pi}{10}\) does not fit this range. Therefore, there are no possible values of \(x\) satisfying the original equation within the range for inverse sine.
Given this, the number of real solutions is:
0
Approach Solution -2
We are given the equation:
\[2 \sin^{-1} x + 3 \cos^{-1} x = \frac{2\pi}{5}\]
Let \(\sin^{-1} x = \alpha\) and \(\cos^{-1} x = \beta\). We know the identity:
\[\sin^{-1} x + \cos^{-1} x = \frac{\pi}{2}\]
So, we have:
\[2\alpha + 3\beta = \frac{2\pi}{5}\]
Using \(\beta = \frac{\pi}{2} - \alpha\), substitute this into the equation:
\[2\alpha + 3\left(\frac{\pi}{2} - \alpha\right) = \frac{2\pi}{5}\]
Simplifying:
\[2\alpha + \frac{3\pi}{2} - 3\alpha = \frac{2\pi}{5}\]
\[-\alpha + \frac{3\pi}{2} = \frac{2\pi}{5}\]
\[-\alpha = \frac{2\pi}{5} - \frac{3\pi}{2}\]
\[-\alpha = \frac{4\pi}{10} - \frac{15\pi}{10} = -\frac{11\pi}{10}\]
\[\alpha = \frac{11\pi}{10}\]
Now, since \(\alpha = \sin^{-1} x\) and \(\sin^{-1} x\) must lie in the range \(\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]\), we find that \(\alpha = \frac{11\pi}{10}\) is not possible because it is outside the allowed range of the inverse sine function.
Thus, the equation has no real solutions.
Final Answer:
0
Top JEE Main Mathematics Questions
- The value of is
- The number of terms of an is even. The sum of the odd terms is and of the even terms is . The last term exceeds the first by . Then the number of terms in the series is ______.
- If , then the general value of is:
- The general solution of
- If the constant term in the binomial expansion of $\left(\frac{x^{\frac{3}{2}}}{2}-\frac{4}{x^l}\right)^9$ is $-84$ and the coefficient of $x^{-3 l}$ is $2^\alpha \beta$, where $\beta<0$ is an odd number, then $|\alpha l-\beta|$ is equal to_____
Top JEE Main Inverse Trigonometric Functions Questions
- If \(a=sin^{-1}(sin\ 5)\) and \(b=cos^{-1}(cos\ 5)\), then \(a^2+b^2=\)
- Considering only the principal values of inverse trigonometric functions, the number of positive real values of \( x \) satisfying \[ \tan^{-1}(x) + \tan^{-1}(2x) = \frac{\pi}{4} \] is:
- If the domain of the function \[\sin^{-1} \left( \frac{3x - 22}{2x - 19} \right) + \log_e \left( \frac{3x^2 - 8x + 5}{x^2 - 3x - 10} \right)\] is \( (\alpha, \beta] \), then \( 3\alpha + 10\beta \) is equal to:
- If the domain of the function \(f(x) = \sin^{-1}\left( \frac{x - 1}{2x + 3} \right)\) is \(\mathbb{R} - (\alpha, \beta)\), then \(12 \alpha \beta\) is equal to:
- Given the inverse trigonometric function assumes principal values only. Let \( x, y \) be any two real numbers in \( [-1, 1] \) such that \[ \cos^{-1}x - \sin^{-1}y = \alpha, \, -\frac{\pi}{2} \leq \alpha \leq \pi. \] Then, the minimum value of \( x^2 + y^2 + 2xy \sin \alpha \) is:
Top JEE Main Questions
- Two masses $m$ and $\frac{m}{2}$ are connected at the two ends of a massless rigid rod of length $l$. The rod is suspended by a thin wire of torsional constant $k$ at the centre of mass of the rod-mass system(see figure). Because of torsional constant $k$, the restoring torque is $\tau =k\theta$ for angular displacement $\theta$. If the rod is rota ted by $\theta_0$ and released, the tension in it when it passes through its mean position will be:
What will be the equilibrium constant of the given reaction carried out in a \(5 \,L\) vessel and having equilibrium amounts of \(A_2\) and \(A\) as \(0.5\) mole and \(2 \times 10^{-6}\) mole respectively?
The reaction : \(A_2 \rightleftharpoons 2A\)- The value of is
- The number of terms of an is even. The sum of the odd terms is and of the even terms is . The last term exceeds the first by . Then the number of terms in the series is ______.
A black body is at a temperature of 2880 K. The energy of radiation emitted by this body with wavelength between 499 nm and 500 nm is U1, between 999 nm and 1000 nm is U2 and between 1499 nm and 1500 nm is U3. The Wien's constant, b = 2.88×106 nm-K. Then,