The number of roots of the equation, (81) sin²x + (81) cos²x = 30 in the interval [ 0, π ] is equal to :

Solution

Maths sir · 3 steps

Since sin² x + cos² x = 1, call one power t, the other is 81/t, and you get a quadratic in t.

1. Substitute. Let t = 81^(sin² x). Because cos² x = 1 - sin² x, the second term is 81^(1-sin² x) = 81/t.

   t + 81/t = 30 ⇒ t² - 30t + 81 = 0 ⇒ t = 27 or t = 3

2. Back to sine. Write everything as a power of 3, using 81 = 3⁴. Then 3^(4 sin² x) = 3³ gives sin² x = 3/4, and 3^(4 sin² x) = 3¹ gives sin² x = 1/4.

   Case	sin x in [0, π]	Roots
   sin² x = 3/4	√3/2 (cannot be negative here)	π/3, 2π/3
   sin² x = 1/4	1/2 (cannot be negative here)	π/6, 5π/6

3. So. On [0, π], sin x ≥ 0, so each case gives exactly two angles. That makes 2 + 2 = 4 roots, which is option (c).

The idea to keep. When the exponents are sin² x and cos² x, they add to 1, so the two powers multiply to a constant. Put t for one of them and solve the quadratic.

Answer from the NTA final answer key. Solution by PrepWiser.