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

Answer: (B)

a2 + 2c - 1 = 0

Answer 2

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

Let's start by understanding the given information: We have a quadratic equation x 2+ax − c =0 and we're told that both sin2 θ and cos2 θ are solutions of this equation.

The quadratic equation can be factored as: x 2+ax − c =(x −sin2 θ)(x −cos2 θ)=0.

Now, expand the factored expression: x 2−(x cos2 θ +x sin2 θ)+(sin2 θ cos2 θ)=0.

Notice that sin2 θ cos2 θ =21​sin4 θ , using the double angle trigonometric identity.

Therefore, the equation becomes: x 2−(x cos2 θ +x sin2 θ)+21​sin4 θ =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 x 2+ax − c =0.

Coefficient of x in the given equation: a Coefficient of x in the expanded equation: −(cos2 θ +sin2 θ)

Setting these coefficients equal: −(cos2 θ +sin2 θ)=a.(1)

Constant term in the given equation: − c Constant term in the expanded equation: 21​sin4 θ

Setting these constant terms equal: (2)21​sin4 θ =− c.(2)

Now, let's square equation (3)(cos2 θ +sin2 θ)2=a 2.(3)

Expanding the left-hand side of equation (3)(3): cos22 θ +2cos2 θ sin2 θ +sin22 θ =a 2.

Using the trigonometric identity sin2 θ +cos2 θ =1: 1+2cos2 θ sin2 θ =a 2. 2(4) 2cos2 θ sin2 θ =a 2−1.(4)

Now, let's substitute equation (2)(2) into equation (4)(4): 2cos2 θ(−sin4 θ 2 c ​)=a 2−1.

Simplifying: −sin4 θ 4 c cos2 θ ​=a 2−1.

Recall the trigonometric identity −4cos⁡2/ 2sin⁡2 cos⁡2=2−1.−2sin2 θ cos2 θ 4 c cos2 θ ​=a 2−1.−sin2 θ 2 c ​=a 2−1. a 2+2 c −1=0.

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