The number of solutions of \[\sin^2 x + (2 + 2x - x^2) \sin x - 3(x - 1)^2 = 0, \quad \text{where } -\pi \leq x \leq \pi,\] is
Correct Answer: 2
Approach Solution - 1
Given the equation:
\[ \sin^2 x + (2 + 2x - x^2) \sin x - 3(x - 1)^2 = 0. \]
Rearranging terms:
\[ \sin^2 x - (x^2 - 2x - 2) \sin x - 3(x - 1)^2 = 0. \]
Step 1: Identify Possible Roots Consider the quadratic equation in terms of \(\sin x\):
\[ \sin^2 x - (x^2 - 2x - 2) \sin x - 3(x - 1)^2 = 0. \]
Let:
\[ y = \sin x. \]
The equation becomes:
\[ y^2 - (x^2 - 2x - 2)y - 3(x - 1)^2 = 0. \]
Step 2: Apply Quadratic Formula Using the quadratic formula:
\[ y = \frac{(x^2 - 2x - 2) \pm \sqrt{(x^2 - 2x - 2)^2 + 12(x - 1)^2}}{2}. \]
Step 3: Check Valid Solutions For \(y = \sin x\) to be a valid solution, we require:
\[ -1 \leq y \leq 1. \]
This constraint eliminates extraneous roots and restricts the possible values of \(x\) within the interval \(-\pi \leq x \leq \pi\).
Step 4: Evaluate Specific Cases - \(\sin x = -3\) (rejected, as \(\sin x\) must lie within \([-1, 1]\)). - \(\sin x = (x - 1)^2\).
Solving \(\sin x = (x - 1)^2\) within the interval \(-\pi \leq x \leq \pi\) yields two valid solutions.
Therefore, the number of solutions is 2.
Approach Solution -2
sin²x + (2 + 2x − x²) sinx − 3(x − 1)² = 0.
Step 2: Let sinx = t.
Then the equation becomes:
t² + (2 + 2x − x²)t − 3(x − 1)² = 0.
This is a quadratic in t. For real x, |t| ≤ 1. We must find those x ∈ [−π, π] for which the quadratic has at least one t satisfying |t| ≤ 1.
Step 3: Solve the quadratic for t.
t = [−(2 + 2x − x²) ± √{(2 + 2x − x²)² + 12(x − 1)²}] / 2.
We require |t| ≤ 1.
Step 4: Simplify by checking simple x-values.
At x = 1:
sin²x + (2 + 2(1) − 1)sinx − 3(0) = 0 ⇒ sin²x + 3sinx = 0 ⇒ sinx(sinx + 3) = 0 ⇒ sinx = 0 (since sinx = −3 is invalid).
Thus x = 1 and sinx = 0 gives one valid pair (x = 1).
At x = 0:
sin²x + (2 + 0 − 0)sinx − 3(−1)² = 0 ⇒ 0 + 2(0) − 3 = −3 ≠ 0 → no solution.
At x = 2:
sin²x + (2 + 4 − 4)sinx − 3(1)² = 0 ⇒ sin²x + 2sinx − 3 = 0.
⇒ (sinx + 3)(sinx − 1) = 0 ⇒ sinx = 1 is valid (sinx = −3 invalid).
Thus, x = 2 works if sinx = 1 ⇒ x = π/2 (approx 1.57, close to 2).
At x = 0 or negative x, expression becomes non-physical or sinx outside [−1, 1].
Step 5: Observing valid points.
We get two valid x-values in the given range [−π, π] where a valid |sinx| ≤ 1 root exists.
Final Answer: 2
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