Solve Trigonometric Equations
Examples and Questions With Solutions
How to solve trigonometric equations? Questions are presented along with detailed solutions and explanations.
How to Solve Trigonometric Functions?
Example 1Find all the solutions of the trigonometric equation √3 sec(θ) + 2 = 0Solution: 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:
.
Example 2Solve the trigonometric equation 2 sin(θ) = -1Solution: 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:
.
Example 3Solve the trigonometric equation √2 cos(3x + π/4) = - 1Solution: 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 4Solve the trigonometric equation - 2 sin^{ 2}x - cos x = - 1Solution: The above equation may be factored if all trigonometric functions included in that equation are the same. So using the identity sin^{ 2}x = 1 - cos^{ 2}x, we can rewrite the above equation using the same trigonometric function cos x as follows: - 2 (1 - cos^{ 2}x) - cos x = - 1 Simplify and rewrite as 2 cos^{ 2}x - 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 x_{1} = 2π/3 + 2nπ , n = 0, ~+mn~ 1 , ~+mn~ 2, ... x_{2} = 4π/3 + 2nπ , n = 0, ~+mn~ 1 , ~+mn~ 2, ... Solve equation (2) cos x = 1 x_{3} = 2nπ , n = 0, ~+mn~ 1 , ~+mn~ 2, ... |
More References and links
Middle School Maths (Grades 6, 7, 8, 9) - Free Questions and Problems With AnswersHigh School Maths (Grades 10, 11 and 12) - Free Questions and Problems With Answers
Primary Maths (Grades 4 and 5) with Free Questions and Problems With Answers
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