Question

If sin 2θ and cos 2θ are solutions of x² + ax - c = 0, then

• A. a² - 2c - 1 = 0
• B. a² + 2c - 1 = 0
• C. a² + 2c + 1 = 0
• D. a² - 2c + 1 = 0

Answer 1

The Correct Option is B

The correct option (B) a² + 2c - 1 = 0.

Let's start by understanding the given information: We have a quadratic equation x2+ax−c=0 and we're told that both sin2θ and cos2θ are solutions of this equation.

The quadratic equation can be factored as: x2+ax−c=(x−sin2θ)(x−cos2θ)=0.

Now, expand the factored expression: x2−(xcos2θ+xsin2θ)+(sin2θcos2θ)=0.

Notice that sin2θcos2θ=21​sin4θ, using the double angle trigonometric identity.

Therefore, the equation becomes: x2−(xcos2θ+xsin2θ)+21​sin4θ=0.

Now, let's compare the coefficients of x and the constant term on the left-hand side of this equation with those in the original equation x2+ax−c=0.

Coefficient of x in the given equation: a Coefficient of x in the expanded equation: −(cos2θ+sin2θ)

Setting these coefficients equal: −(cos2θ+sin2θ)=a.(1)

Constant term in the given equation: −c Constant term in the expanded equation: 21​sin4θ

Setting these constant terms equal: (2)21​sin4θ=−c.(2)

Now, let's square equation (3)(cos2θ+sin2θ)2=a2.(3)

Expanding the left-hand side of equation (3)(3): cos22θ+2cos2θsin2θ+sin22θ=a2.

Using the trigonometric identity sin2θ+cos2θ=1: 1+2cos2θsin2θ=a2. 2(4) 2cos2θsin2θ=a2−1.(4)

Now, let's substitute equation (2)(2) into equation (4)(4): 2cos2θ(−sin4θ2c​)=a2−1.

Simplifying: −sin4θ4ccos2θ​=a2−1.

Recall the trigonometric identity −4cos⁡2/ 2sin⁡2 cos⁡2=2−1.−2sin2θcos2θ4ccos2θ​=a2−1.−sin2θ2c​=a2−1. a2+2c−1=0.

Thus, we have shown that a2+2c−1=0, which justifies the given answer.