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Question: If $2sin^{-1}x = -sin^{-1}\left(2x\sqrt{1-x^2}\right) - \pi$, then $x$ satisfies...

If 2sin−1x=−sin−1(2x1−x2)−π2sin^{-1}x = -sin^{-1}\left(2x\sqrt{1-x^2}\right) - \pi, then xx satisfies

A

12≤x≤1\frac{1}{\sqrt{2}} \leq x \leq 1

B

0≤x≤120 \leq x \leq \frac{1}{\sqrt{2}}

C

−1≤x≤−12-1 \leq x \leq -\frac{1}{\sqrt{2}}

D

−12≤x≤0-\frac{1}{\sqrt{2}} \leq x \leq 0

Answer

−1≤x≤−12-1 \leq x \leq -\frac{1}{\sqrt{2}}

Explanation

Solution

To solve the equation 2sin⁡−1x=−sin⁡−1(2x1−x2)−π2\sin^{-1}x = -\sin^{-1}\left(2x\sqrt{1-x^2}\right) - \pi, we follow these steps:

  1. Determine the domain of xx:
    For sin⁡−1x\sin^{-1}x to be defined, we must have −1≤x≤1-1 \leq x \leq 1.
    For sin⁡−1(2x1−x2)\sin^{-1}\left(2x\sqrt{1-x^2}\right) to be defined, we must have −1≤2x1−x2≤1-1 \leq 2x\sqrt{1-x^2} \leq 1.

  2. Substitute x=sin⁡θx = \sin\theta:
    Let x=sin⁡θx = \sin\theta. Since −1≤x≤1-1 \leq x \leq 1, we can choose θ∈[−π2,π2]\theta \in \left[-\frac{\pi}{2}, \frac{\pi}{2}\right].
    The left side of the equation becomes 2sin⁡−1(sin⁡θ)=2θ2\sin^{-1}(\sin\theta) = 2\theta.

    The term 2x1−x22x\sqrt{1-x^2} can be simplified:
    2sin⁡θ1−sin⁡2θ=2sin⁡θcos⁡2θ2\sin\theta\sqrt{1-\sin^2\theta} = 2\sin\theta\sqrt{\cos^2\theta}.
    Since θ∈[−π2,π2]\theta \in \left[-\frac{\pi}{2}, \frac{\pi}{2}\right], cos⁡θ≥0\cos\theta \geq 0.
    So, cos⁡2θ=cos⁡θ\sqrt{\cos^2\theta} = \cos\theta.
    Therefore, 2x1−x2=2sin⁡θcos⁡θ=sin⁡(2θ)2x\sqrt{1-x^2} = 2\sin\theta\cos\theta = \sin(2\theta).

  3. Rewrite the equation in terms of θ\theta:
    The original equation becomes:
    2θ=−sin⁡−1(sin⁡(2θ))−π2\theta = -\sin^{-1}(\sin(2\theta)) - \pi.

  4. Analyze sin⁡−1(sin⁡(2θ))\sin^{-1}(\sin(2\theta)) based on the range of 2θ2\theta:
    Since θ∈[−π2,π2]\theta \in \left[-\frac{\pi}{2}, \frac{\pi}{2}\right], 2θ∈[−π,π]2\theta \in [-\pi, \pi].
    We need to consider the definition of sin⁡−1(sin⁡y)\sin^{-1}(\sin y) for y∈[−π,π]y \in [-\pi, \pi]:

    • If y∈[−π2,π2]y \in \left[-\frac{\pi}{2}, \frac{\pi}{2}\right], then sin⁡−1(sin⁡y)=y\sin^{-1}(\sin y) = y.
    • If y∈(π2,π]y \in \left(\frac{\pi}{2}, \pi\right], then sin⁡−1(sin⁡y)=π−y\sin^{-1}(\sin y) = \pi - y.
    • If y∈[−π,−π2)y \in \left[-\pi, -\frac{\pi}{2}\right), then sin⁡−1(sin⁡y)=−π−y\sin^{-1}(\sin y) = -\pi - y.

    Let's apply these cases to y=2θy = 2\theta:

    Case 1: −π2≤2θ≤π2-\frac{\pi}{2} \leq 2\theta \leq \frac{\pi}{2}
    This implies −π4≤θ≤π4-\frac{\pi}{4} \leq \theta \leq \frac{\pi}{4}.
    In this interval, sin⁡−1(sin⁡(2θ))=2θ\sin^{-1}(\sin(2\theta)) = 2\theta.
    The equation becomes:
    2θ=−(2θ)−π2\theta = -(2\theta) - \pi
    4θ=−π4\theta = -\pi
    θ=−π4\theta = -\frac{\pi}{4}.
    This value θ=−π4\theta = -\frac{\pi}{4} lies within the interval [−π4,π4]\left[-\frac{\pi}{4}, \frac{\pi}{4}\right]. So, it is a valid solution for θ\theta.

    Case 2: π2<2θ≤π\frac{\pi}{2} < 2\theta \leq \pi
    This implies π4<θ≤π2\frac{\pi}{4} < \theta \leq \frac{\pi}{2}.
    In this interval, sin⁡−1(sin⁡(2θ))=π−2θ\sin^{-1}(\sin(2\theta)) = \pi - 2\theta.
    The equation becomes:
    2θ=−(π−2θ)−π2\theta = -(\pi - 2\theta) - \pi
    2θ=−π+2θ−π2\theta = -\pi + 2\theta - \pi
    0=−2π0 = -2\pi.
    This is a contradiction, so there are no solutions in this interval.

    Case 3: −π≤2θ<−π2-\pi \leq 2\theta < -\frac{\pi}{2}
    This implies −π2≤θ<−π4-\frac{\pi}{2} \leq \theta < -\frac{\pi}{4}. (Note: The boundary 2θ=−π22\theta = -\frac{\pi}{2} i.e., θ=−π4\theta = -\frac{\pi}{4} is included in Case 1, but for continuous functions, it's often included in both boundary cases. Let's check if it creates an issue. If θ=−π4\theta = -\frac{\pi}{4}, 2θ=−π22\theta = -\frac{\pi}{2}, sin⁡−1(sin⁡(−π2))=−π2\sin^{-1}(\sin(-\frac{\pi}{2})) = -\frac{\pi}{2}. Using the formula −π−y-\pi - y, we get −π−(−π2)=−π2-\pi - (-\frac{\pi}{2}) = -\frac{\pi}{2}. So the formula holds for 2θ=−π22\theta = -\frac{\pi}{2} as well. Thus, we can consider the interval −π2≤θ≤−π4-\frac{\pi}{2} \leq \theta \leq -\frac{\pi}{4}.)
    In this interval, sin⁡−1(sin⁡(2θ))=−π−2θ\sin^{-1}(\sin(2\theta)) = -\pi - 2\theta.
    The equation becomes:
    2θ=−(−π−2θ)−π2\theta = -(-\pi - 2\theta) - \pi
    2θ=π+2θ−π2\theta = \pi + 2\theta - \pi
    2θ=2θ2\theta = 2\theta.
    This is an identity, which means all values of θ\theta in the interval [−π2,−π4]\left[-\frac{\pi}{2}, -\frac{\pi}{4}\right] are solutions.

  5. Combine the solutions for θ\theta:
    From Case 1, θ=−π4\theta = -\frac{\pi}{4} is a solution.
    From Case 3, all θ∈[−π2,−π4]\theta \in \left[-\frac{\pi}{2}, -\frac{\pi}{4}\right] are solutions.
    Combining these, the set of solutions for θ\theta is [−π2,−π4]\left[-\frac{\pi}{2}, -\frac{\pi}{4}\right].

  6. Convert back to xx:
    We have x=sin⁡θx = \sin\theta. Since sin⁡θ\sin\theta is an increasing function in the interval [−π2,π2]\left[-\frac{\pi}{2}, \frac{\pi}{2}\right], we can apply sin⁡\sin to the interval for θ\theta:
    sin⁡(−π2)≤sin⁡θ≤sin⁡(−π4)\sin\left(-\frac{\pi}{2}\right) \leq \sin\theta \leq \sin\left(-\frac{\pi}{4}\right)
    −1≤x≤−12-1 \leq x \leq -\frac{1}{\sqrt{2}}.

The range of xx that satisfies the given equation is −1≤x≤−12-1 \leq x \leq -\frac{1}{\sqrt{2}}.