Class 12 Inverse Trigonometric Functions · Short Answer2 Marksinverse_trigonometric_equations

Class 12 Maths · Inverse Trigonometric Functions · Short Answer #62

Solve \(\sin^{-1}(4x)+\sin^{-1}(3x)=-\pi/2\).

✅ Step-by-Step Solution

Final Answer: x=−1/5

The common domain is [−1/4,1/4]. The negative right side forces x<0, since both arcsines have the sign of x.
Rearrange arcsin4x=−π/2−arcsin3x and take sine: 4x=−√(1−9x²). Squaring gives 16x²=1−9x², so x²=1/25; the sign leaves x=−1/5.
For x=−1/5, the positive acute angles arcsin(4/5) and arcsin(3/5) are complementary; their negatives sum to −π/2. Thus x=−1/5 is the unique solution.

[0.5 marks] The common domain is [−1/4,1/4]. The negative right side forces x<0, since both arcsines have the sign of x.
[1 mark] Rearrange arcsin4x=−π/2−arcsin3x and take sine: 4x=−√(1−9x²). Squaring gives 16x²=1−9x², so x²=1/25; the sign leaves x=−1/5.
[0.5 marks] For x=−1/5, the positive acute angles arcsin(4/5) and arcsin(3/5) are complementary; their negatives sum to −π/2. Thus x=−1/5 is the unique solution.

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