Solve Trigonometric Equations
Tutorial, with detailed explanations, on how to solve trigonometric equations using different methods and strategies and the properties of trigonometric functions and identities. The unit circle helps in locating the solutions once you have the reference angle.
Many more examples on how to solve trigonomteric equations with detailed solutions - Grade 12, are presented in this site.
Examples with Solutions
Example 1Solve the trigonometric equation.(find all solutions)
Solution to example 1
solve for cos(x)
cos x = 1/2
solve for x by finding all values in the
interval [0 , 2pi) that satisfy the above trigonometric equation.
In this case, with cosine positive and equal to 1 / 2, there
are two values: one in the first quadrant of the unit circle.
x1 = pi / 3
and a second one in the fourth quadrant (see the two solutions in unit circle in figure below).
x2 = 2*pi - pi / 3 = 5*pi / 3
find all solutions using the fact that of cos x has a period of 2pi
x1 = pi / 3 + 2*k*pi
x2 = 5*pi / 3 + 2*k*pi
where k is any integer
conclusion: There is an infinite number of solutions which can be generated by giving different values to k.
Example 2Find the solutions in the interval [0 , 2pi) for the trigonometric equation
Solution to example 2
change cos 2x to 1 - sin 2x
-5(1 - sin 2x) + 9 sin x = -3
multiply factors and group to obtain
5 sin2x + 9 sin x -2 = 0
let u = sinx and substitute to obtain an
5 u2 + 9 u - 2 = 0
use any method to solve for u.
By the quadratic formula, we obtain two solutions u1 and u2
u1 = [ -9 - sqrt(121) ] / 10 = 1 / 5 = -2
u2 = [ -9 + sqrt(121) ] / 10 = 0.2
we now solve the equation for x
u1 = sin x = -2
the above equation has no solutions for x since -2 is not in the range of values of sin(x);
-1 <= sin x <= 1.
u2 = sin x = 0.2
sin x is positive in the first and second
quadrants of the unit circle. There are two
solutions to the above equation in [0 , 2 pi)(see unit circle below)
solution in the first quadrant
x1 = arcsin 0.2
arcsin 0.2 is the inverse sine function
and it can be approximated using a calculator)
solution in the second quadrant
x2 = pi - arcsin 0.2
conclusion: There are two solutions to the given equation.
Example 3Find all solutions for the trigonometric equation
Solution to example 3
subtract cot x from both sides of the equation and simplify
cot x cos 2x - cot x = 0
Factor cot x
cot x (cos 2x - 1) = 0
Setting each factor in the above trigonometric equation to zero, we obtain two equations.
cot x = 0 and cos 2x - 1 = 0
The solutions to equation cot x = 0 are given by
x = pi / 2 + k*pi , k is am integer.
- Equation cos 2x - 1 = 0 gives
cos x = 1 and cos x = -1
The solutions to the above equations are given by
x = 2k*pi and x = (2k + 1)*pi where k is an integer
HOWEVER the above cannot be solution to the given equation since cot x is undefined for x = 2k*pi and x = (2k + 1)*pi.
conclusion: The solutions to the given equation are.
x = pi / 2 + k*pi where k is an integer.
Note that many of the techniques used in solving algebraic equations are also used to solve trigonometric equations.