What is the first step in solving √3 cos 2θ + 8 cos θ + 3√3 = 0?

The first step is to use the double angle identity cos 2θ = 2 cos² θ - 1 to convert the equation into a quadratic form in terms of cos θ.

How do we solve the quadratic equation 2√3 cos² θ + 8 cos θ + 2√3 = 0?

We can use the quadratic formula or middle-term splitting. Factoring gives (√3 cos θ + 1)(2 cos θ + 2√3) = 0, or simply (√3 cos θ + 3)(2√3 cos θ + 2) = 0.

What are the possible values for cos θ in this problem?

The algebraic solutions are cos θ = -1/√3 and cos θ = -√3.

Why is cos θ = -√3 rejected?

The value of cos θ must always lie in the range [-1, 1]. Since -√3 is approximately -1.732, it is outside this range and has no real solution.

How many solutions does cos θ = -1/√3 have in one period [0, 2π]?

In the interval [0, 2π], cos θ is negative in the 2nd and 3rd quadrants, providing exactly 2 solutions.

How do we count solutions in the interval [-3π, 2π]?

The total length of the interval is 5π. We count the number of times the graph of cos θ intersects the line y = -1/√3 in this specific range.

What is the value of θ where cos θ = -1/√3?

θ = π ± arccos(1/√3) + 2nπ. These are specific points in the second and third quadrants.

Does the endpoint -3π or 2π include a solution?

At θ = -3π and θ = 2π, cos θ is -1 and 1 respectively. Neither equals -1/√3, so the endpoints are not solutions.

Is there a graphical way to find the number of solutions?

Yes, by drawing y = cos θ and the horizontal line y = -1/√3 from -3π to 2π, we can count the intersection points.

What is the final answer for the number of solutions?

There are a total of 5 solutions in the given interval [-3π, 2π].

Number of solutions of √3 cos 2θ + 8 cos θ + 3√3 = 0, θ ∈ [-3π, 2π] is :

Number of solutions of √3 cos 2θ + 8 cos θ + 3√3 = 0, θ ∈ [-3π, 2π] is : | JEE Main Mathematics

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Math Trigonometry

Q MCQ Trig Equations

Number of solutions of $\sqrt{3} \cos 2\theta + 8 \cos \theta + 3\sqrt{3} = 0, \theta \in [-3\pi, 2\pi]$ is :

A) $5$ B) $4$ C) $3$ D) $0$

✅ Correct Answer

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Solution Steps

Simplify the Equation

Given equation: $\sqrt{3} \cos 2\theta + 8 \cos \theta + 3\sqrt{3} = 0$

Using the identity $\cos 2\theta = 2 \cos^2 \theta – 1$:

$\sqrt{3}(2 \cos^2 \theta – 1) + 8 \cos \theta + 3\sqrt{3} = 0$

$2\sqrt{3} \cos^2 \theta + 8 \cos \theta + 2\sqrt{3} = 0$

Solve the Quadratic in $\cos \theta$

Let $t = \cos \theta$. The equation becomes $2\sqrt{3}t^2 + 8t + 2\sqrt{3} = 0$.

Dividing by 2: $\sqrt{3}t^2 + 4t + \sqrt{3} = 0$.

Using middle term splitting: $\sqrt{3}t^2 + 3t + t + \sqrt{3} = 0$.

$\sqrt{3}t(t + \sqrt{3}) + 1(t + \sqrt{3}) = 0 \implies (\sqrt{3}t + 1)(t + \sqrt{3}) = 0$.

Determine Valid Values of $\cos \theta$

Case 1: $t = -\sqrt{3} \implies \cos \theta = -\sqrt{3} \approx -1.732$. Rejected (since $-1 \le \cos \theta \le 1$).

Case 2: $t = -1/\sqrt{3} \implies \cos \theta = -1/\sqrt{3}$. This is valid as $|-1/\sqrt{3}| < 1$.

Analyze Interval $[-3\pi, 2\pi]$

We need to find the number of solutions for $\cos \theta = -1/\sqrt{3}$ in $[-3\pi, 2\pi]$.

Interval break-down:

1. $[-3\pi, -\pi]$: $2$ solutions (one in 2nd quadrant of rotation, one in 3rd).

2. $[-\pi, \pi]$: $2$ solutions (at $\pm (\pi – \alpha)$ where $\cos \alpha = 1/\sqrt{3}$).

3. $[\pi, 2\pi]$: $1$ solution (in the second quadrant of the cycle $[0, 2\pi]$, specifically in $(\pi, 3\pi/2)$).

Counting Intersections

Interval length is $5\pi = 2.5$ periods. The value $-1/\sqrt{3}$ is negative.

In $[-3\pi, -2\pi]$: $1$ solution.

In $[-2\pi, 0]$: $2$ solutions.

In $[0, 2\pi]$: $2$ solutions.

Total Solutions = 1 + 2 + 2 = 5 (Option A)

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Theory

1. Trigonometric Double Angle Identities

The identity $\cos 2\theta = 2 \cos^2 \theta – 1$ is fundamental for solving equations that mix linear and double-angle terms. By substituting this identity, we transform a trigonometric equation into a polynomial (quadratic) equation in terms of a single trigonometric ratio. This is a primary strategy in JEE Mathematics to reduce complexity. Other forms include $1 – 2 \sin^2 \theta$ or $\cos^2 \theta – \sin^2 \theta$, but the choice depends on matching the other terms in the equation.

2. Range of Trigonometric Functions

Both $\sin \theta$ and $\cos \theta$ are bounded functions, with a range of $[-1, 1]$. Any algebraic solution for a trigonometric variable that falls outside this interval (like $-\sqrt{3}$ in this problem) must be discarded as it represents no real solution in the complex plane for real angles. Recognizing these bounds early saves significant time during competitive exams and prevents the inclusion of extraneous solutions.

3. Periodicity and Counting Solutions

Cosine is a periodic function with a period of $2\pi$. In a full cycle $[0, 2\pi]$, the equation $\cos \theta = k$ (where $|k| < 1$ and $k \neq \pm 1$) has exactly two solutions. For negative $k$, these solutions lie in the 2nd and 3rd quadrants. When the interval is extended (like $[-3\pi, 2\pi]$), we must count how many times the function cycles through those quadrants. Sketching the cosine curve and drawing a horizontal line at $y = k$ is the most reliable visual method for counting solutions accurately.

4. General Solutions vs. Specific Intervals

While the general solution for $\cos \theta = \cos \alpha$ is $\theta = 2n\pi \pm \alpha$, JEE Main often asks for the number of solutions in a constrained interval. This requires evaluating the general formula for various integer values of $n$ to check which results fall within the given range. For the interval $[-3\pi, 2\pi]$, we check $n = -1, 0, 1$. Mastery over both algebraic general solutions and interval testing is essential for high-scoring performance in calculus and trigonometry.

5. Middle Term Splitting for Irrationals

Quadratic equations with irrational coefficients, such as $\sqrt{3}t^2 + 4t + \sqrt{3} = 0$, can be solved by recognizing factors that sum to the linear coefficient. Here, $3$ and $1$ are factors of $(\sqrt{3} \times \sqrt{3}) = 3$ that sum to $4$. This technique is faster than the quadratic formula in many cases. Practice with such radical coefficients is common in JEE preparations to improve algebraic manipulation speed and accuracy.

6. Quadrant Analysis for Negative Ratios

The sign of a trigonometric ratio determines the quadrant of the angle. For $\cos \theta = -1/\sqrt{3}$, the negative sign indicates that the solutions must be in the 2nd quadrant ($\pi/2$ to $\pi$) and the 3rd quadrant ($\pi$ to $3\pi/2$). By mapping the given range $[-3\pi, 2\pi]$ onto these quadrants, we can verify the count. In this problem, the range covers two full cycles plus a half-cycle, leading to the final tally of 5 intersection points.

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FAQs

Why did we use $2 \cos^2 \theta – 1$ and not $1 – 2 \sin^2 \theta$?

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The original equation contained a $8 \cos \theta$ term. To get a quadratic in a single variable, we needed to convert $\cos 2\theta$ into $\cos \theta$.

Is $-1/\sqrt{3}$ a standard value for cosine?

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No, it is not a standard angle like 30° or 45°. However, we don’t need the exact angle, only to know that it is a valid value between -1 and 1.

How exactly did we get 5 solutions?

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Interval $[-3\pi, -2\pi]$ has 1 solution, $[-2\pi, 0]$ has 2 solutions, and $[0, 2\pi]$ has 2 solutions. Total = 1 + 2 + 2 = 5.

What would happen if the interval was open $(-3\pi, 2\pi)$?

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Since neither $-3\pi$ nor $2\pi$ are solutions (as $\cos(-3\pi) = -1$ and $\cos(2\pi) = 1$), the count would still be 5.

Why is $\cos \theta = -\sqrt{3}$ not possible?

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The square root of 3 is approximately 1.732. The cosine function never goes below -1, so -1.732 is impossible.

Can I solve this by squaring both sides?

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Squaring usually introduces extraneous solutions and makes the equation harder (degree 4). The substitution method is much safer.

What is the periodicity of the total function?

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The function is a combination of $\cos 2\theta$ and $\cos \theta$, so the overall period is the LCM of $\pi$ and $2\pi$, which is $2\pi$.

How do quadrants help in counting solutions?

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Cosine is negative in Q2 and Q3. By checking how many times the interval passes through these quadrants, we find the number of solutions.

What if the question asked for the sum of solutions?

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Then you would use the general solution formula and sum the specific values within the range.

Is this a common JEE Main topic?

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Yes, “Number of solutions” for trigonometric equations is one of the most frequently asked topics in JEE Main Mathematics.

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