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Trigonometric Functions and Equations question

2025 · 2 Apr · Shift 1 · Q29
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  5. /2025 · 2 Apr · Shift 1 · Q29

Trigonometric Functions and Equations question

2025 · 2 Apr · Shift 1 · Q29

JEE MainMathematicsTrigonometric Functions and EquationsMCQ+4 / −1
If θ∈[−2π,2π]\theta \in[-2 \pi, 2 \pi]θ∈[−2π,2π], then the number of solutions of 22cos⁡2θ+(2−6)cos⁡θ−3=02 \sqrt{2} \cos ^2 \theta+(2-\sqrt{6}) \cos \theta-\sqrt{3}=022​cos2θ+(2−6​)cosθ−3​=0, is equal to:
  1. A
    8
  2. B
    6
  3. C
    10
  4. D
    12
View written solutionFree

Correct answer: A

  1. Let x=cos⁡θx=\cos\thetax=cosθ. Then the equation becomes

22x2+(2−6)x−3=0.2\sqrt{2}x^2+(2-\sqrt{6})x-\sqrt{3}=0.22​x2+(2−6​)x−3​=0.

We must first solve this quadratic for xxx, and then count how many θ∈[−2π,2π]\theta\in[-2\pi,2\pi]θ∈[−2π,2π] satisfy cos⁡θ=x\cos\theta=xcosθ=x.

  1. Factor the quadratic.

We look for a factorization of the form

22x2+(2−6)x−3=(ax+b)(cx+d).2\sqrt{2}x^2+(2-\sqrt{6})x-\sqrt{3}=(ax+b)(cx+d).22​x2+(2−6​)x−3​=(ax+b)(cx+d).

Try

(2x+3)(2x−1).(\sqrt{2}x+\sqrt{3})(2x-1).(2​x+3​)(2x−1).

Expanding:

(2x+3)(2x−1)=22x2−2x+23x−3.(\sqrt{2}x+\sqrt{3})(2x-1) =2\sqrt{2}x^2-\sqrt{2}x+2\sqrt{3}x-\sqrt{3}.(2​x+3​)(2x−1)=22​x2−2​x+23​x−3​.

Now,

−2+23=2−6?-\sqrt{2}+2\sqrt{3}=2-\sqrt{6}?−2​+23​=2−6​?

Let us instead try

(2x−3)(2x+1).(\sqrt{2}x-\sqrt{3})(2x+1).(2​x−3​)(2x+1).

Expanding:

(2x−3)(2x+1)=22x2+2x−23x−3(\sqrt{2}x-\sqrt{3})(2x+1) =2\sqrt{2}x^2+\sqrt{2}x-2\sqrt{3}x-\sqrt{3}(2​x−3​)(2x+1)=22​x2+2​x−23​x−3​

so the coefficient of xxx is

2−23.\sqrt{2}-2\sqrt{3}.2​−23​.

This is not equal to 2−62-\sqrt{6}2−6​ directly. So we solve using the quadratic formula.

  1. Use the quadratic formula for

22x2+(2−6)x−3=0.2\sqrt{2}x^2+(2-\sqrt{6})x-\sqrt{3}=0.22​x2+(2−6​)x−3​=0.

Here,

a=22,b=2−6,c=−3.a=2\sqrt{2},\quad b=2-\sqrt{6},\quad c=-\sqrt{3}.a=22​,b=2−6​,c=−3​.

Then

x=−(2−6)±(2−6)2−4(22)(−3)2⋅22.x=\frac{-(2-\sqrt{6})\pm\sqrt{(2-\sqrt{6})^2-4(2\sqrt{2})(-\sqrt{3})}}{2\cdot 2\sqrt{2}}.x=2⋅22​−(2−6​)±(2−6​)2−4(22​)(−3​)​​.

Compute the discriminant:

(2−6)2+86=4+6−46+86=10+46.(2-\sqrt{6})^2+8\sqrt{6} =4+6-4\sqrt{6}+8\sqrt{6} =10+4\sqrt{6}.(2−6​)2+86​=4+6−46​+86​=10+46​.

Notice that

10+46=(2+6)2.10+4\sqrt{6}=(2+\sqrt{6})^2.10+46​=(2+6​)2.

So

10+46=2+6.\sqrt{10+4\sqrt{6}}=2+\sqrt{6}.10+46​​=2+6​.

Hence

x=−2+6±(2+6)42.x=\frac{-2+\sqrt{6}\pm(2+\sqrt{6})}{4\sqrt{2}}.x=42​−2+6​±(2+6​)​.

Thus the two roots are:

  • With +++ sign: x=−2+6+2+642=2642=32.x=\frac{-2+\sqrt{6}+2+\sqrt{6}}{4\sqrt{2}}=\frac{2\sqrt{6}}{4\sqrt{2}}=\frac{\sqrt{3}}{2}.x=42​−2+6​+2+6​​=42​26​​=23​​.

  • With −-− sign: x=−2+6−2−642=−442=−12=−22.x=\frac{-2+\sqrt{6}-2-\sqrt{6}}{4\sqrt{2}}=\frac{-4}{4\sqrt{2}}=-\frac{1}{\sqrt{2}}=-\frac{\sqrt{2}}{2}.x=42​−2+6​−2−6​​=42​−4​=−2​1​=−22​​.

So the equation reduces to

cos⁡θ=32orcos⁡θ=−22.\cos\theta=\frac{\sqrt{3}}{2} \quad \text{or} \quad \cos\theta=-\frac{\sqrt{2}}{2}.cosθ=23​​orcosθ=−22​​.

  1. Count solutions for cos⁡θ=32\cos\theta=\frac{\sqrt{3}}{2}cosθ=23​​ in [−2π,2π][-2\pi,2\pi][−2π,2π].

General angles are

θ=2kπ±π6.\theta=2k\pi\pm\frac{\pi}{6}.θ=2kπ±6π​.

List those in [−2π,2π][-2\pi,2\pi][−2π,2π]:

−11π6, −π6, π6, 11π6.-\frac{11\pi}{6},\ -\frac{\pi}{6},\ \frac{\pi}{6},\ \frac{11\pi}{6}.−611π​, −6π​, 6π​, 611π​.

So there are 4 solutions.

  1. Count solutions for cos⁡θ=−22\cos\theta=-\frac{\sqrt{2}}{2}cosθ=−22​​ in [−2π,2π][-2\pi,2\pi][−2π,2π].

General angles are

θ=2kπ±3π4.\theta=2k\pi\pm\frac{3\pi}{4}.θ=2kπ±43π​.

List those in [−2π,2π][-2\pi,2\pi][−2π,2π]:

−5π4, −3π4, 3π4, 5π4.-\frac{5\pi}{4},\ -\frac{3\pi}{4},\ \frac{3\pi}{4},\ \frac{5\pi}{4}.−45π​, −43π​, 43π​, 45π​.

So there are 4 solutions.

  1. Total number of solutions:

4+4=8.4+4=8.4+4=8.

Therefore, the correct option is

A: 8.\boxed{\text{A: }8}.A: 8​.

  1. Comparison with stored answer:

Stored correct answer is A, which matches our derived answer.

Next

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