Trigonometric Functions: Exercise 3.4 - General Solutions

Understanding Trigonometric Equations and Their Solutions

A trigonometric equation is an equation that involves trigonometric functions of unknown angles. For example, sin x = 1/2 or 2cos²x + 3sin x = 0 are trigonometric equations. Unlike linear or quadratic equations which often have a fixed number of solutions, trigonometric equations can have infinitely many solutions due to the periodic nature of sine, cosine, and tangent functions. For instance, if sin x = 1/2, then x can be π/6, 5π/6, 2π + π/6, 2π + 5π/6, -2π + π/6, and so on.

To manage this, we categorize solutions into two types:

1. Principal Solutions: These are the solutions that lie within the interval [0, 2π). This interval covers one full cycle of the trigonometric functions, so any principal solution will be unique within this range for a given value.
2. General Solutions: These are expressions that represent all possible solutions to a trigonometric equation. They involve an arbitrary integer n (where n ∈ Z), reflecting the periodicity. For example, if sin x = sin y, the general solution for x will relate to y and nπ to capture all coterminal angles and reflections. Understanding how to derive and apply these general solution formulas is the core objective of Exercise 3.4.

Essential General Solution Formulas

If sin x = sin y

The general solution is x = nπ + (-1)^n y, where n ∈ Z.

If cos x = cos y

The general solution is x = 2nπ ± y, where n ∈ Z.

If tan x = tan y

The general solution is x = nπ + y, where n ∈ Z.

If sin²x = sin²y or cos²x = cos²y or tan²x = tan²y

The general solution for all three cases is x = nπ ± y, where n ∈ Z. This is a handy consolidated formula for squared trigonometric equations.

Step-by-Step Approach to Solving Trigonometric Equations

1. Step 1: Simplify the Equation — If necessary, use trigonometric identities to simplify the equation, express it in terms of a single trigonometric function, or factorize it. The goal is to bring it to one of the standard forms like sin x = k, cos x = k, or tan x = k (where k is a constant), or a product of such forms.
2. Step 2: Find Principal Solutions (if required) — Determine the principal values of the angle y such that sin y = k, cos y = k, or tan y = k. Remember that principal solutions lie in [0, 2π). If the problem asks for general solutions directly, you just need one suitable value for y (often the smallest positive one, or from [-π/2, π/2] for sine/tan and [0, π] for cosine).
3. Step 3: Apply the General Solution Formula — Once you have the equation in a standard form (e.g., sin x = sin y), apply the corresponding general solution formula. Ensure you correctly identify y from your principal solutions or a chosen reference angle.
4. Step 4: State the Integer 'n' — Always include n ∈ Z (where Z denotes the set of all integers) as part of your final general solution. This signifies that n can be any positive, negative, or zero integer, leading to the infinite solutions.
5. Step 5: Verify (Optional but Recommended) — Substitute a few values of n (e.g., n=0, 1, -1) into your general solution and check if the resulting angles satisfy the original equation. This helps catch errors.

Worked Examples: Finding General Solutions

• Example 1: Find the general solution for sin x = -√3/2 Step 1: Identify the reference angle. We know sin(π/3) = √3/2. So, the reference angle is π/3. Step 2: Determine quadrants for negative sine. Sine is negative in the 3rd and 4th quadrants. Step 3: Find a value for 'y' (principal solution or suitable reference). In the 3rd quadrant, y = π + π/3 = 4π/3. In the 4th quadrant, y = 2π - π/3 = 5π/3 (or -π/3). Let's choose y = -π/3 (as sin(-π/3) = -sin(π/3) = -√3/2). Step 4: Apply the general solution formula for sin x = sin y. The formula is x = nπ + (-1)^n y, where n ∈ Z. Substitute y = -π/3: x = nπ + (-1)^n (-π/3) x = nπ - (-1)^n (π/3) Final Answer: The general solution is x = nπ - (-1)^n (π/3), where n ∈ Z. (Alternatively, using y = 4π/3, the solution would be x = nπ + (-1)^n (4π/3) which is also correct.)
• Example 2: Solve cos 2x = cos 4x for the general solution. Step 1: Rearrange the equation. We have cos 2x = cos 4x. This is directly in the form cos A = cos B (where A = 2x and B = 4x). Step 2: Apply the general solution formula for cos A = cos B. The formula is A = 2nπ ± B, where n ∈ Z. So, 2x = 2nπ ± 4x. Step 3: Solve for x for both positive and negative cases. Case 1: 2x = 2nπ + 4x 2x - 4x = 2nπ -2x = 2nπ x = -nπ Since n can be any integer, -n also covers all integers. So, x = nπ where n ∈ Z. Case 2: 2x = 2nπ - 4x 2x + 4x = 2nπ 6x = 2nπ x = (2nπ)/6 x = nπ/3 where n ∈ Z. Final Answer: The general solutions are x = nπ and x = nπ/3, where n ∈ Z.
• Example 3: Find the general solution of tan x = cot x. Step 1: Express both sides in terms of the same trigonometric function. We know that cot x = 1/tan x or cot x = tan(π/2 - x). Let's use cot x = tan(π/2 - x). The equation becomes tan x = tan(π/2 - x). Step 2: Apply the general solution formula for tan A = tan B. The formula is A = nπ + B, where n ∈ Z. Here, A = x and B = π/2 - x. So, x = nπ + (π/2 - x). Step 3: Solve for x. x + x = nπ + π/2 2x = nπ + π/2 2x = π(n + 1/2) x = (π/2)(n + 1/2) x = (π/2) * ((2n+1)/2) x = (2n+1)π/4 Final Answer: The general solution is x = (2n+1)π/4, where n ∈ Z.

Exam Tips & Common Mistakes in Trigonometric Equations

Solving trigonometric equations requires careful application of identities and formulas. Here are crucial tips to avoid common pitfalls:

• Do NOT divide by variable trigonometric functions: A common mistake is to divide both sides of an equation by sin x or cos x. For example, in sin x cos x = cos x, dividing by cos x would lead to sin x = 1, making you lose the solutions where cos x = 0. Instead, always factorize: sin x cos x - cos x = 0 => cos x (sin x - 1) = 0. This gives cos x = 0 OR sin x = 1.
• Remember n ∈ Z: Always explicitly state 'where n ∈ Z' (n belongs to the set of integers) in your general solutions. Omitting this can lead to loss of marks.
• Simplifying y for general solution: When sin x = k, you need a y such that sin y = k. It's often easiest to pick the smallest positive angle or the angle in the principal value branch (e.g., [-π/2, π/2] for sine, [0, π] for cosine, (-π/2, π/2) for tangent) to represent y in the general solution formula. However, any correct y will yield a valid (though perhaps differently structured) general solution.
• Check the domain: Some problems might specify a particular domain (e.g., 0 ≤ x < 2π). In such cases, after finding the general solution, substitute integer values for n to find solutions within the specified domain.
• Identities are your friends: Many complex equations can be simplified using fundamental identities (e.g., sin²x + cos²x = 1, double angle formulas, sum-to-product formulas). Practice recalling and applying them efficiently.

Practice Questions with Solutions

• Q: Find the general solution of tan 2x = -cot(x + π/3). A: Step 1: Convert cot to tan using cot θ = tan(π/2 - θ). tan 2x = -tan(π/2 - (x + π/3)) tan 2x = -tan(π/2 - x - π/3) tan 2x = -tan(π/6 - x) Step 2: Use the identity -tan θ = tan(-θ). tan 2x = tan(-(π/6 - x)) tan 2x = tan(x - π/6) Step 3: Apply the general solution formula tan A = tan B => A = nπ + B. 2x = nπ + x - π/6 Step 4: Solve for x. 2x - x = nπ - π/6 x = nπ - π/6 Final answer: x = nπ - π/6, where n ∈ Z.
• Q: Find the general solution for sin 2x - sin 4x + sin 6x = 0. A: Step 1: Group terms and apply sum-to-product identity: sin A + sin B = 2sin((A+B)/2)cos((A-B)/2). (sin 6x + sin 2x) - sin 4x = 0 2sin((6x+2x)/2)cos((6x-2x)/2) - sin 4x = 0 2sin(4x)cos(2x) - sin 4x = 0 Step 2: Factor out sin 4x. sin 4x (2cos 2x - 1) = 0 Step 3: Set each factor to zero and solve. Case 1: sin 4x = 0 Using sin θ = 0 => θ = nπ. 4x = nπ x = nπ/4, where n ∈ Z. Case 2: 2cos 2x - 1 = 0 2cos 2x = 1 cos 2x = 1/2 We know cos(π/3) = 1/2. So, cos 2x = cos(π/3). Using cos θ = cos α => θ = 2mπ ± α. 2x = 2mπ ± π/3 x = mπ ± π/6, where m ∈ Z. Final answer: x = nπ/4 and x = mπ ± π/6, where n, m ∈ Z.
• Q: Determine the general solution of sec² 2x = 1 - tan 2x. A: Step 1: Use the identity sec²θ = 1 + tan²θ. 1 + tan² 2x = 1 - tan 2x Step 2: Rearrange the equation to form a quadratic equation in terms of tan 2x. tan² 2x + tan 2x = 0 Step 3: Factorize the equation. tan 2x (tan 2x + 1) = 0 Step 4: Set each factor to zero and solve. Case 1: tan 2x = 0 Using tan θ = 0 => θ = nπ. 2x = nπ x = nπ/2, where n ∈ Z. Case 2: tan 2x + 1 = 0 tan 2x = -1 We know tan(π/4) = 1. Since tan is negative in the 2nd and 4th quadrants, we can use tan(-π/4) = -1 or tan(3π/4) = -1. Let's use tan 2x = tan(-π/4). Using tan θ = tan α => θ = mπ + α. 2x = mπ - π/4 x = mπ/2 - π/8, where m ∈ Z. Final answer: x = nπ/2 and x = mπ/2 - π/8, where n, m ∈ Z.
• Q: Find the general solution for cos x + cos 3x = 0. A: Step 1: Use the sum-to-product identity: cos A + cos B = 2cos((A+B)/2)cos((A-B)/2). 2cos((x+3x)/2)cos((x-3x)/2) = 0 2cos(2x)cos(-x) = 0 2cos(2x)cos(x) = 0 (since cos(-x) = cos x) Step 2: Set each factor to zero and solve. Case 1: cos x = 0 Using cos θ = 0 => θ = (2n+1)π/2. x = (2n+1)π/2, where n ∈ Z. Case 2: cos 2x = 0 Using cos θ = 0 => θ = (2m+1)π/2. 2x = (2m+1)π/2 x = (2m+1)π/4, where m ∈ Z. Final answer: x = (2n+1)π/2 and x = (2m+1)π/4, where n, m ∈ Z.

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