Question

If the equation \( \sin^4 x - (p + 2)\sin^2 x - (p + 3) = 0 \) has a solution, then \( p \) must lie in the interval:

• A. \( [-3, -2] \)
• B. \( (-3, -2) \)
• C. \( (2, 3) \)
• D. \( [-5, -3] \)

Hint:

When dealing with trigonometric equations involving powers, substituting \( \sin^2 x = t \) (or \( \cos^2 x = t \)) simplifies the problem into solving a quadratic within a bounded domain \( [0,1] \).

Answer 1

The Correct Option is A

Step 1: Let \( \sin^2 x = t \).
Substituting, we get the quadratic in \( t \): \[ t^2 - (p+2)t - (p+3) = 0 \] where \( 0 \leq t \leq 1 \) since \( t = \sin^2 x \).
Step 2: Solve the quadratic equation.
Using the quadratic formula: \[ t = \frac{(p+2) \pm \sqrt{(p+2)^2 + 4(p+3)}}{2} \] Simplify the discriminant: \[ (p+2)^2 + 4(p+3) = p^2 + 4p + 4 + 4p + 12 = p^2 + 8p + 16 \] Thus: \[ t = \frac{(p+2) \pm \sqrt{p^2 + 8p + 16}}{2} \]
Step 3: Condition for solution in \([0, 1]\).
For a solution, at least one root \( t \) must lie in \([0,1]\). Analyze critical points:
Let’s check when \( t=0 \) or \( t=1 \).
Substitute \( t=0 \) into the quadratic:
\[ 0 - 0 - (p+3) = 0 \quad \Rightarrow \quad p = -3 \] Substitute \( t=1 \) into the quadratic: \[ 1 - (p+2)(1) - (p+3) = 0 \quad \Rightarrow \quad 1 - p -2 - p -3 = 0 \] \[ -2p -4 = 0 \quad \Rightarrow \quad p = -2 \] Thus, \( p \) must lie between \(-3\) and \(-2\), including endpoints. Hence, \( p \in [-3, -2] \).