Question

If \(2\sin x - \cos 2x = 1\), then \( (3 - 2\sin^2x) = \)

• A. \( \sqrt{3} \)
• B. \( -\sqrt{3} \)
• C. \( \sqrt{5} \)
• D. \( -\sqrt{5} \) Correct Answer

Hint:

When solving trigonometric equations, use identities to express the equation in terms of a single trigonometric function if possible. For quadratic equations in \( \sin x \) or \( \cos x \), solve for the trigonometric function and ensure the values lie within their valid range ([-1, 1]).

Answer 1

The Correct Option is C

Step 1: Use the identity \( \cos 2x = 1 - 2\sin^2 x \) in the given equation.
\[ 2\sin x - (1 - 2\sin^2 x) = 1 \] \[ 2\sin x - 1 + 2\sin^2 x = 1 \] \[ 2\sin^2 x + 2\sin x - 2 = 0 \] \[ \sin^2 x + \sin x - 1 = 0 \]
Step 2: Let \( y = \sin x \).
The equation becomes \( y^2 + y - 1 = 0 \).
Using the quadratic formula, \( y = \frac{-1 \pm \sqrt{1^2 - 4(1)(-1)}}{2(1)} = \frac{-1 \pm \sqrt{1+4}}{2} = \frac{-1 \pm \sqrt{5}}{2} \).
Since \( -1 \le \sin x \le 1 \), we must choose the appropriate value.
\( \frac{-1+\sqrt{5}}{2} \approx \frac{-1+2.
236}{2} = \frac{1.
236}{2} = 0.
618 \) (Valid).
\( \frac{-1-\sqrt{5}}{2} \approx \frac{-1-2.
236}{2} = \frac{-3.
236}{2} = -1.
618 \) (Not valid).
So, \( \sin x = \frac{-1+\sqrt{5}}{2} \).

Step 3: We need to find the value of \( (3 - 2\sin^2x) \).
From \( \sin^2 x + \sin x - 1 = 0 \), we have \( \sin^2 x = 1 - \sin x \).
Substitute this into the expression: \[ 3 - 2\sin^2x = 3 - 2(1 - \sin x) = 3 - 2 + 2\sin x = 1 + 2\sin x \]
Step 4: Substitute the value of \( \sin x \).
\[ 1 + 2\sin x = 1 + 2\left(\frac{-1+\sqrt{5}}{2}\right) = 1 + (-1+\sqrt{5}) = 1 - 1 + \sqrt{5} = \sqrt{5} \] This matches option (3).