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Question

If $\sin(3x + 10^{\circ}) = \cos(2x - 5^{\circ})$, find x.

This question was previously asked in
SSC CGL 2025 Tier 2 Paper 1 Question Paper (19-Jan-2026)
The correct answer is
$17^{\circ}$

Solving Trigonometric Equation for x

The problem requires finding the value of x given the trigonometric equation: $ \sin(3x + 10^{\circ}) = \cos(2x - 5^{\circ}) $

Using Complementary Angle Identity

We use the trigonometric identity relating sine and cosine: $ \sin(\theta) = \cos(90^{\circ} - \theta) $ Applying this to the given equation, let $ \theta = 3x + 10^{\circ} $. Therefore, $ \sin(3x + 10^{\circ}) = \cos(90^{\circ} - (3x + 10^{\circ})) $

Equating Cosine Terms

Now, substitute this back into the original equation: $ \cos(90^{\circ} - (3x + 10^{\circ})) = \cos(2x - 5^{\circ}) $ For the cosine values to be equal, their arguments must be equal (considering the general solution, we take the simplest case): $ 90^{\circ} - (3x + 10^{\circ}) = 2x - 5^{\circ} $

Step-by-Step Calculation for x

  1. Simplify the left side: $ 90^{\circ} - 3x - 10^{\circ} = 2x - 5^{\circ} $ $ 80^{\circ} - 3x = 2x - 5^{\circ} $
  2. Rearrange the terms to group x terms and constant terms: $ 80^{\circ} + 5^{\circ} = 2x + 3x $
  3. Combine like terms: $ 85^{\circ} = 5x $
  4. Solve for x by dividing both sides by 5: $ x = \frac{85^{\circ}}{5} $ $ x = 17^{\circ} $

Thus, the value of x is $17^{\circ}$.

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