# Solve Trigonometric Equations - Examples and Questions With Detailed Solutions

How to solve trigonometric equations? Grade 11 trigonometry questions are presented along with detailed solutions and explanations.

 How to Solve Trigonometric Functions? Example 1: Find all the solutions of the trigonometric equation √3 sec(θ) + 2 = 0 Solution: Using the identity sec(θ) = 1 / cos(θ), we rewrite the equation in the form cos(θ) = - √3 / 2 Find the reference θr angle by solving cos(θr) = √3 / 2 for θr acute. θr = π/6 Use the reference angle θr to determine the solutions θ1 and θ2 on the interval [0 , 2π) of the given equation. The equation cos(θ) = - √3 / 2 suggests that cos(θ) is negative and that means the terminal side of angle θ solution to the given equation is either in quadrants II or III as shown below using the unit circle. Hence the solutions: θ1 = π - θr = π - π/6 = 5π/6 θ2 = π + θr = π + π/6 = 7π/6 Use the solutions on the interval [0 , 2π) to find all solutions by adding multiples of 2π as follows: θ1 = 5π/6 + 2nπ , n = 0, ~+mn~ 1 , ~+mn~ 2, ... θ2 = 7π/6 + 2nπ , n = 0, ~+mn~ 1 , ~+mn~ 2, ... Below are shown the graphical solutions on the interval [0 , 2π) . Example 2: Solve the trigonometric equation 2 sin(θ) = -1 Solution: Rewrite the above equation in simple form as shown below. sin(θ) = -1/2 Find the reference θr angle by solving sin(θ) = 1/2 for θr acute. θr = π/6 Use the reference angle θr to determine the solutions θ1 and θ2 on the interval [0 , 2π) of the given equation. The equation sin(θ) = - 1 / 2 suggests that sin(θ) is negative and that means the terminal side of angle θ is either in quadrants III or VI as shown in the unit circle below. Hence the solutions: θ1 = π + θr = 7π/6 θ2 = 2π - θr = 11π/6 Use the solutions on the interval [0 , 2π) to find all solutions by adding multiples of 2π as follows: θ1 = 7π/6 + 2nπ , n = 0, ~+mn~ 1 , ~+mn~ 2, ... θ2 = 11π/6 + 2nπ , n = 0, ~+mn~ 1 , ~+mn~ 2, ... . Example 3: Solve the trigonometric equation √2 cos(3x + π/4) = - 1 Solution: Let θ = 3x + π/4 and rewrite the equation in simple form. √2 cos(θ) = - 1 cos(θ) = -1/√2 Find the reference θr angle by solving cos(θ) = 1/√2 for θr acute. θr = π/4 Use the reference angle θr to determine the solutions θ1 and θ2 on the interval [0 , 2π) of the given equation. The equation cos(θ) = - 1/√2 suggests that cos(θ) is negative and that means the terminal side of angle θ is either in quadrants II or III. Hence the two solutions of the equation cos(θ) = - 1/√2 on the interval [0 , 2π) are given by θ1 = π - θr = 3π/4 θ2 = π + θr = 5π/4 We now write the general solutions by adding multiples of 2π as follows: θ1 = 3π/4 + 2nπ , n = 0, ~+mn~ 1 , ~+mn~ 2, ... θ2 = 5π/4 + 2nπ , n = 0, ~+mn~ 1 , ~+mn~ 2, ... We now substitute θ1 and θ2 by the expression 3x + π/4 3x + π/4 = 3π/4 + 2nπ 3x + π/4 = 5π/4 + 2nπ and solve for x to obtain the solutions for x. x = π/6 + 2nπ/3 , n = 0, ~+mn~ 1 , ~+mn~ 2, ... x = π/3 + 2nπ/3 , n = 0, ~+mn~ 1 , ~+mn~ 2, ... Example 4: Solve the trigonometric equation - 2 sin 2x - cos x = - 1 Solution: The above equation may be factored if all trigonometric functions included in that equation are the same. So using the identity sin 2x = 1 - cos 2x, we can rewrite the above equation using the same trigonometric function cos x as follows: - 2 (1 - cos 2x) - cos x = - 1 Simplify and rewrite as 2 cos 2x - cos x - 1 = 0 Factor the left hand side (2 cos x + 1)(cos x - 1) = 0 Hence the two equations to solve (1) 2 cos x + 1 = 0 and (2) cos x - 1 = 0 Solve equation (1) using the reference angle as was done in the examples above. cos x = -1/2 x1 = 2π/3 + 2nπ , n = 0, ~+mn~ 1 , ~+mn~ 2, ... x2 = 4π/3 + 2nπ , n = 0, ~+mn~ 1 , ~+mn~ 2, ... Solve equation (2) cos x = 1 x3 = 2nπ , n = 0, ~+mn~ 1 , ~+mn~ 2, ...

Updated: 20 January 2017 (A Dendane)