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📐 Mathematics  ·  Inverse Trigonometric Functions  ·  JEE

Solve: sin-1(x) + sin-1(2x) = pi/3. Approximate the smaller positive root region check, given x must satisfy domain constraints.

Answer: x = sqrt(3)/(2sqrt(7)) is the value satisfying domain and equation.

  • A x = 1/(2sqrt(7)), obtained by dropping the sqrt(3) factor in the derivation
  • B x = sqrt(3)/(2sqrt(7)) is the value satisfying domain and equation
  • C x = 1/2, which fails the domain constraint required for sin-1(2x)
  • D x = 1/sqrt(7), obtained from an algebra slip in clearing the radical

Correct answer: B. x = sqrt(3)/(2sqrt(7)) is the value satisfying domain and equation

Explanation: Setting sin-1(2x) = pi/3 - sin-1(x) and taking sine of both sides with the addition formula leads, after squaring and simplifying (7x<sup>2</sup> = 3/4 form), to x = sqrt(3)/(2sqrt(7)), which satisfies both the equation and domain |2x|<=1.

y = sin⁻¹x: Principal Value Branchxyx=-1x=1y=-π/2y=π/2Domain restricted to [-1,1]; range restricted to [-π/2,π/2] - this restriction is what makes the inverse exist

The graph of sin⁻¹x is confined to a narrow domain [-1,1] (since sine itself only takes values in that range) and range [-π/2,π/2] (the principal value branch chosen to make sine one-one and invertible there).

Concept context

Restricting trig functions to make them invertible, the principal value branches of sin-inverse, cos-inverse, tan-inverse, and friends, and the key identities relating them.

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