Notes On Trigonometric Equations - CBSE Class 11 Maths
Equations: A mathematical statement that shows equality between two expressions is called an equation. Trigonometric Equations: An equation involving trigonometric functions of a variable is known as a trigonometric equation. Example: a cos θ + b sin θ = 0, p tan2 θ + q sec2 θ + r = 0 sin x = 1/2 Solution of a trigonometric equation: The value of the unknown angle that satisfies a given trigonometric equation. Example: sin x = 0 ⇒ x = 0, π, 2π,...… 2. cos x = 0 ⇒ x = $\frac{\pi }{\text{2}}$, 3$\frac{\pi }{\text{2}}$,5$\frac{\pi }{\text{2}}$...… The solution of a trigonometric equation, for which 0 ≤ x < 2π, are called the principal solutions. 0 and π are the principal solutions of sin x = 0. π/2 and 3π/2 are the principal solutions of cos x = 0. Find the principal solutions of tan x = -√3. But tan $\frac{\pi }{\text{3}}$ = √3. Also, the tan function is negative in the second and the fourth quadrants. tan(π - $\frac{\pi }{\text{3}}$) = -tan $\frac{\pi }{\text{3}}$ = -√3 and tan(2π - $\frac{\pi }{\text{3}}$) = - tan $\frac{\pi }{\text{3}}$ = -√3 i.e. tan 2$\frac{\pi }{\text{3}}$ = tan 5$\frac{\pi }{\text{3}}$ = -√3 Hence, 2$\frac{\pi }{\text{3}}$ and 5$\frac{\pi }{\text{3}}$ are the principal solutions Ex: Find the principal solutions of sin x = 1/2. But sin $\frac{\pi }{\text{6}}$ = ½. Also, sin(π - $\frac{\pi }{\text{6}}$) = sin 5$\frac{\pi }{\text{6}}$ = ½. Hence, $\frac{\pi }{\text{6}}$ and 5$\frac{\pi }{\text{6}}$ are the principal solutions. Trigonometric functions are periodic functions. The functions sin, cos, cosec and sec repeat after an interval of 2π, while the functions tan and cot repeat after an interval of π. sin q = 0 ⇒ q = nπ, n ∈ Z When the solution set of a trigonometric equation is an expression involving an integer n, the solution is called the general solution of the trigonometric equation. cos q = 0 ⇒ q = (2n+1)$\frac{\pi }{\text{2}}$, n ∈ Z Theorem For any real numbers x and y, sin x = sin y implies x = nπ + (-1)ny, where n ∈ Z . Proof For any real numbers x and y, sin x = sin y. ⇒sin x - sin y = 0 ⇒ 2cos ($\frac{\text{x + y}}{\text{2}}$). sin ($\frac{\text{x - y}}{\text{2}}$) = 0/2 = 0 ⇒ Either cos ($\frac{\text{x + y}}{\text{2}}$) = 0 or sin ($\frac{\text{x - y}}{\text{2}}$) = 0 sin q = 0 ⇒ q =nπ and, cos q = 0 ⇒ q = (2n+1)π/2,n ∈ Z Therefore, ($\frac{\text{x + y}}{\text{2}}$) = (2n+1)π/2 or ($\frac{\text{x - y}}{\text{2}}$) = np, where n ∈ Z i.e. x= (2n + 1)π - y or x = 2nπ + y, where n ∈ Z (-1)2n+1 = -1 and (-1)2n = 1, where n ∈ Z Hence, x = (2n + 1) p + (-1)2n+1y or x = 2nπ + (-1)2ny, where n ∈ Z. Combining these two results, we get x = nπ + (-1)ny, where n ∈ Z. Theorem For any real numbers x and y, cos x = cos y, implies x = 2nπ±y, where n ∈ Z Proof For any real numbers x and y, cos x = cos y. ⇒ cos x - cos y = 0. ⇒ -2 sin ((x+y)/2). sin ((x-y)/2) = 0 ⇒ sin ((x+y)/2). sin ((x-y)/2) = 0/2 = 0 ⇒ sin ((x+y)/2) = 0 or sin ((x-y)/2) = 0 sin q = 0 ⇒ q = nπ, n ∈ Z Therefore, ((x+y)/2) = np or ((x-y)/2) = n p, where n ∈ Z i.e. x = 2nπ-y or x = 2nπ+y where n ∈ Z Hence, x = 2nπ ± y where n ∈ Z. Theorem If x and y are not odd multiples of π/2, then tan x = tan y implies x = np+ y, where n ∈ Z. Proof Suppose x and y are not odd multiples of π/2 and tan x = tan y. ⇒ tan x - tan y = 0 ⇒ sin x / cos x - sin y / cos y = 0 ⇒ (sin x cos y - cos x sin y)/ cos x cos y = 0 ⇒sin(x - y)/cos x cos y = 0 ⇒ sin(x- y) = 0 sin θ = 0 ⇒ θ = nπ, n ∈ Z Therefore, x - y = nπ i.e. x = nπ + y, where n ∈ Z Solution set of Trigonometric Functions The solution set of sin θ = k, k ∈ R and -1 ≤ k ≤ 1, is {nπ + (-1)ny/n ∈ Z}, where y is the principal solution. The solution set of cos θ = k, k ∈ R and -1 ≤ k ≤ 1,  {2nπ + y/n ∈ Z} where y is the principal solution. The solution set of tan θ = k, k ∈ R is {nπ + y/n ∈ Z}, where y is the principal solution.

#### Summary

Equations: A mathematical statement that shows equality between two expressions is called an equation. Trigonometric Equations: An equation involving trigonometric functions of a variable is known as a trigonometric equation. Example: a cos θ + b sin θ = 0, p tan2 θ + q sec2 θ + r = 0 sin x = 1/2 Solution of a trigonometric equation: The value of the unknown angle that satisfies a given trigonometric equation. Example: sin x = 0 ⇒ x = 0, π, 2π,...… 2. cos x = 0 ⇒ x = $\frac{\pi }{\text{2}}$, 3$\frac{\pi }{\text{2}}$,5$\frac{\pi }{\text{2}}$...… The solution of a trigonometric equation, for which 0 ≤ x < 2π, are called the principal solutions. 0 and π are the principal solutions of sin x = 0. π/2 and 3π/2 are the principal solutions of cos x = 0. Find the principal solutions of tan x = -√3. But tan $\frac{\pi }{\text{3}}$ = √3. Also, the tan function is negative in the second and the fourth quadrants. tan(π - $\frac{\pi }{\text{3}}$) = -tan $\frac{\pi }{\text{3}}$ = -√3 and tan(2π - $\frac{\pi }{\text{3}}$) = - tan $\frac{\pi }{\text{3}}$ = -√3 i.e. tan 2$\frac{\pi }{\text{3}}$ = tan 5$\frac{\pi }{\text{3}}$ = -√3 Hence, 2$\frac{\pi }{\text{3}}$ and 5$\frac{\pi }{\text{3}}$ are the principal solutions Ex: Find the principal solutions of sin x = 1/2. But sin $\frac{\pi }{\text{6}}$ = ½. Also, sin(π - $\frac{\pi }{\text{6}}$) = sin 5$\frac{\pi }{\text{6}}$ = ½. Hence, $\frac{\pi }{\text{6}}$ and 5$\frac{\pi }{\text{6}}$ are the principal solutions. Trigonometric functions are periodic functions. The functions sin, cos, cosec and sec repeat after an interval of 2π, while the functions tan and cot repeat after an interval of π. sin q = 0 ⇒ q = nπ, n ∈ Z When the solution set of a trigonometric equation is an expression involving an integer n, the solution is called the general solution of the trigonometric equation. cos q = 0 ⇒ q = (2n+1)$\frac{\pi }{\text{2}}$, n ∈ Z Theorem For any real numbers x and y, sin x = sin y implies x = nπ + (-1)ny, where n ∈ Z . Proof For any real numbers x and y, sin x = sin y. ⇒sin x - sin y = 0 ⇒ 2cos ($\frac{\text{x + y}}{\text{2}}$). sin ($\frac{\text{x - y}}{\text{2}}$) = 0/2 = 0 ⇒ Either cos ($\frac{\text{x + y}}{\text{2}}$) = 0 or sin ($\frac{\text{x - y}}{\text{2}}$) = 0 sin q = 0 ⇒ q =nπ and, cos q = 0 ⇒ q = (2n+1)π/2,n ∈ Z Therefore, ($\frac{\text{x + y}}{\text{2}}$) = (2n+1)π/2 or ($\frac{\text{x - y}}{\text{2}}$) = np, where n ∈ Z i.e. x= (2n + 1)π - y or x = 2nπ + y, where n ∈ Z (-1)2n+1 = -1 and (-1)2n = 1, where n ∈ Z Hence, x = (2n + 1) p + (-1)2n+1y or x = 2nπ + (-1)2ny, where n ∈ Z. Combining these two results, we get x = nπ + (-1)ny, where n ∈ Z. Theorem For any real numbers x and y, cos x = cos y, implies x = 2nπ±y, where n ∈ Z Proof For any real numbers x and y, cos x = cos y. ⇒ cos x - cos y = 0. ⇒ -2 sin ((x+y)/2). sin ((x-y)/2) = 0 ⇒ sin ((x+y)/2). sin ((x-y)/2) = 0/2 = 0 ⇒ sin ((x+y)/2) = 0 or sin ((x-y)/2) = 0 sin q = 0 ⇒ q = nπ, n ∈ Z Therefore, ((x+y)/2) = np or ((x-y)/2) = n p, where n ∈ Z i.e. x = 2nπ-y or x = 2nπ+y where n ∈ Z Hence, x = 2nπ ± y where n ∈ Z. Theorem If x and y are not odd multiples of π/2, then tan x = tan y implies x = np+ y, where n ∈ Z. Proof Suppose x and y are not odd multiples of π/2 and tan x = tan y. ⇒ tan x - tan y = 0 ⇒ sin x / cos x - sin y / cos y = 0 ⇒ (sin x cos y - cos x sin y)/ cos x cos y = 0 ⇒sin(x - y)/cos x cos y = 0 ⇒ sin(x- y) = 0 sin θ = 0 ⇒ θ = nπ, n ∈ Z Therefore, x - y = nπ i.e. x = nπ + y, where n ∈ Z Solution set of Trigonometric Functions The solution set of sin θ = k, k ∈ R and -1 ≤ k ≤ 1, is {nπ + (-1)ny/n ∈ Z}, where y is the principal solution. The solution set of cos θ = k, k ∈ R and -1 ≤ k ≤ 1,  {2nπ + y/n ∈ Z} where y is the principal solution. The solution set of tan θ = k, k ∈ R is {nπ + y/n ∈ Z}, where y is the principal solution.

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