Given that sin (5x-28)° = cos (3x-50)", 0°≤ x ≤ 90°, find the value of x.

Analyze the question: We have sin(5x − 28)° = cos(3x − 50)°, and we need x between 0° and 90°.

Concept: sin θ = cos(90° − θ), so if sin A = cos B, then A + B = 90° (for angles in the acute range).

Solution:

(5x − 28) + (3x − 50) = 90 8x − 78 = 90 8x = 168 x = 21

Check:

• 5(21) − 28 = 105 − 28 = 77°, so sin 77°
• 3(21) − 50 = 63 − 50 = 13°, so cos 13°
• sin 77° = cos(90° − 77°) = cos 13° ✓

Other possibilities: Cosine can also be equal for angles related by A = 90° − B + 360°n or sin A = sin(180° − A), which gives other equations:

• sin A = cos B means A = 90° − B or A = 90° + B (since cos B = cos(−B), sin(90°+B) = cos B).
• Using A = 90° + B: 5x − 28 = 90 + 3x − 50 gives 2x = 68, x = 34. Check: sin(142°) = sin 38° ≈ 0.616, and cos(52°) ≈ 0.616. This also works mathematically, but 34 is not among the options.

Answer: C (21)

Tip: When you see sin = cos, immediately think "complementary angles": add the two angles and set the sum equal to 90°. It saves a lot of time in the exam. Then check your answer against the options.