Problem 2 - Comprehensive Trigonometry for IIT JEE Main and Advanced Rejaul Makshud MeGraw Hill

Question

Solve the following equations and tick the correct one. If \(3 \tan ^{2} \theta-2 \sin \theta=0\), then \(\theta\) is (a) \(n \pi\) (b) \(n \pi+(-1)^{n} \frac{\pi}{6}\) (c) \(n \pi-(-1)^{n} \frac{\pi}{6}\) (d) \(n \pi+\frac{\pi}{3}\)

Step-by-Step Solution

Answer

After doing all necessary steps, it is concluded that the correct option is (b) \(n \pi+(-1)^{n} \frac{\pi}{6}\). Short answer must be a very short answer to the question based on the solution steps.

Step 1 Simplification

Firstly, simplify the equation \(3 \tan ^{2} \theta-2 \sin \theta = 0\) by substituting the trigonometric identity \(\tan ^{2} \theta = \frac{\sin ^{2} \theta}{\cos ^{2} \theta}\), you should get \(\frac{3\sin ^{2} \theta}{\cos ^{2} \theta}-2 \sin \theta = 0\).

Step 2 Remove Fraction

Multiply the equation by \(\cos ^{2} \theta\) to eliminate the fraction. This results in: \(3\sin ^{2} \theta-2 \sin \theta \cos ^{2} \theta = 0\).

Step 3 Factorization

You can factor out \(\sin \theta\) from the equation. Doing so gives \(\sin \theta(3 \sin \theta -2 \cos ^{2} \theta) = 0\).

Step 4 Solving the Equation

Setting each factor equal to zero gives two cases. \(\sin \theta=0\) or \(\sin \theta = \frac{2 \cos ^{2} \theta}{3}\). The solution for \(\sin \theta=0\) is \(\theta=n \pi\). For the second case, substitute \(\cos ^{2} \theta= 1-\sin ^{2} \theta\) and solve for \(\sin \theta\), you should get \(\sin \theta=\frac{1}{2}\). The solutions for this case are \(\theta=n \pi+(-1)^{n} \frac{\pi}{6}\).

Key Concepts

Trigonometric Identities

Trigonometric identities are essential for transforming and simplifying equations in mathematics. They often express one trigonometric function in terms of others.
For example, in the problem, we used the identity \(\tan ^{2} \theta = \frac{\sin ^{2} \theta}{\cos ^{2} \theta}\). Using this identity allows us to transform the expression and move away from tangent, which can be complex to work with.
Remembering these identities can be very helpful:

• \(\tan ^{2}\theta = \frac{\sin ^{2}\theta}{\cos ^{2}\theta}\)
• \(\sin^{2}\theta + \cos^{2}\theta = 1\)
• \(1 + \tan^{2}\theta = \sec^{2}\theta\)

Mastering these identities, you can solve trigonometric equations more easily because they're like shortcuts that lead you directly to your goal.

Factorization

Factorization simplifies equations by breaking them down into simpler factors. In the exercise, after multiplying to remove fractions, we factored the expression \(3\sin^{2}\theta - 2\sin\theta\cos^{2}\theta = 0\). This was simplified by factoring out \(\sin \theta\), resulting in \(\sin \theta (3 \sin \theta - 2 \cos^{2} \theta) = 0\).
Factorization helps by reducing the complexity of a problem, making it easier to find solutions.
Key steps in factorization:

• Identify common elements that can be taken out as a factor.
• Break the expression into parts that multiply back to the original equation.
• This often leads to multiple simpler equations, each of which can be solved independently.

In this case, we identified that either \(\sin \theta = 0\) or \(3 \sin \theta = 2 \cos^{2} \theta\), both providing straightforward paths to solutions.

Solutions of Trigonometric Equations

Solving trigonometric equations often involves finding all possible angles \(\theta\) that satisfy a given equation. This requires a strong understanding of both trigonometric identities and factorization.
In this problem, after factorizing the equation, we have two simpler equations:
1. \(\sin \theta = 0\), which simplifies to solutions \(\theta = n\pi\), where \(n\) is an integer reflecting the periodicity of the sine function.
2. \(\sin \theta = \frac{1}{2}\) was derived from the equation \(3 \sin \theta = 2 \cos^{2} \theta\), giving solutions \(\theta = n\pi + (-1)^n \frac{\pi}{6}\).
Trigonometric equations can have infinite solutions due to their periodic nature, and recognizing this is crucial when listing solutions.

IIT JEE Advanced Mathematics

IIT JEE Advanced Mathematics is one of the most challenging exams, requiring deep understanding and application of mathematical concepts, especially in trigonometry. Problems like the one we solved are typical and require mastering several math skills simultaneously.
This particular problem put to test:

• Comprehension of trigonometric identities, which simplify complex trigonometric functions.
• The ability to apply factorization to break down a complex equation into manageable parts.
• Solving equations effectively, recognizing the general solutions for trigonometric equations.

Preparation for such exams demands practice with a broad set of problems, sharpening both algebraic manipulation and trigonometric theory.