Class 11 RS Aggarwal Chapter-17 Trigonometric Equations Solutions - Free PDF Download
FAQs on RS Aggarwal Class 11 Solutions Chapter-17 Trigonometric Equations
1. How do I find the principal solution of a trigonometric equation using the method in RS Aggarwal?
To find the principal solution for a trigonometric equation, you need to find the value of the angle (θ) that lies within the range of 0 ≤ θ < 2π. The process is as follows:
First, solve the equation to find the reference angle, which is the acute angle corresponding to the positive value of the trigonometric ratio.
Next, identify the quadrants where the solution can lie based on the sign (positive or negative) of the trigonometric function in the original equation.
Finally, calculate the angle in those specific quadrants to find all values within the [0, 2π) interval. These values are the principal solutions.
2. What is the step-by-step process for finding the general solution for an equation like sin x = k in RS Aggarwal Class 11?
The step-by-step process for finding the general solution for sin x = k (where |k| ≤ 1) is:
Step 1: Find the principal value, let's call it α, such that sin α = k and α is in the range [-π/2, π/2].
Step 2: Apply the standard formula for the general solution of sine functions, which is x = nπ + (-1)ⁿα, where 'n' is any integer (n ∈ Z).
This single formula covers all possible solutions because the term (-1)ⁿ correctly adjusts the angle for both even and odd values of 'n', placing the solution in the correct quadrant.
3. How are problems involving cos x = cos α solved in Chapter 17 of RS Aggarwal?
To solve an equation in the form cos x = cos α, you first identify the value of α from the given equation. Then, you apply the standard formula for the general solution of cosine functions, which is x = 2nπ ± α, where 'n' is any integer (n ∈ Z). This formula accounts for the periodic and even nature of the cosine function, covering all possible solutions.
4. What is the correct method for solving trigonometric equations in RS Aggarwal that need to be factorised first?
The correct method for solving trigonometric equations that can be factorised involves these steps:
Rearrange: Move all terms to one side of the equation, making the other side equal to zero.
Factorise: Factor the trigonometric expression. This often involves techniques like taking out a common factor or using quadratic factorisation (e.g., treating sin x as a variable).
Equate to Zero: Set each factor equal to zero independently.
Solve: Solve each of the resulting simpler trigonometric equations to find their respective general solutions. The final answer is the set of all solutions obtained from each factor.
5. Why is finding the general solution crucial for trigonometric equations, beyond just the principal solution?
Finding the general solution is crucial because trigonometric functions are periodic, meaning their values repeat at regular intervals. A principal solution only provides answers within a single cycle (e.g., [0, 2π)). The general solution provides a comprehensive formula that represents all possible values for the angle across the entire real number line. This is essential for applications where the process is continuous or cyclical, such as in physics (wave mechanics) and engineering (signal processing).
6. What is a common mistake to avoid when solving equations by squaring both sides, for instance, in questions from RS Aggarwal Chapter 17?
The most common mistake when solving a trigonometric equation by squaring both sides is the introduction of extraneous roots. Squaring can make negative terms positive, creating solutions that do not satisfy the original equation. To avoid this error, you must verify every solution by substituting it back into the initial, unsquared equation. Any solution that does not satisfy the original equation must be discarded.
7. How do I approach solving equations of the form a cos x + b sin x = c as given in RS Aggarwal solutions?
The most efficient method to solve equations of the form a cos x + b sin x = c is to convert the expression on the left into a single cosine or sine function. Here is the step-by-step approach:
Divide the entire equation by √(a²+b²).
The equation becomes: (a/√(a²+b²))cos x + (b/√(a²+b²))sin x = c/√(a²+b²).
Let a/√(a²+b²) = cos α and b/√(a²+b²) = sin α. The equation simplifies to cos(x - α) = c/√(a²+b²).
Now, solve this standard cosine equation to find the general solution for (x - α), and then solve for x.
8. Why is the solution for tan x = 0 different from sin x = 0, even though both are zero when the numerator is zero?
While both tan x and sin x are zero at the same points (x = nπ), the reasoning is subtly different. For sin x = 0, the solution is simply where the y-coordinate on the unit circle is zero, which is at 0, π, 2π, etc., leading to the general solution x = nπ. For tan x = sin x / cos x = 0, the condition is not only that sin x = 0 but also that cos x ≠ 0, to avoid division by zero. Since cos x is non-zero at x = nπ, the solutions are valid and identical. However, understanding this distinction is key for problems where tan x is undefined, which is not a concern for sin x.
