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.
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Examples with Solutions
Example 1Solve the trigonometric equation.(find all solutions) 2 cos x + 2 = 3
Solution to example 1solve for cos(x)cos x = 1/2solve for x by finding all values in theinterval <0 , 2pi) that satisfy the above trigonometric equation.In this case, with cosine positive and equal to 1 / 2, thereare two values: one in the first quadrant of the unit circle.x1 = pi / 3and 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 2pix1 = pi / 3 + 2*k*pix2 = 5*pi / 3 + 2*k*piwhere k is any integerconclusion: There is an infinite number of solutions whichcan be generated by giving different values to k.
Example 2Find the solutions in the interval <0 , 2pi) for the trigonometric equation-5 cos 2x + 9 sin x = -3
Solution to example 2change cos 2x to 1 - sin 2x-5(1 - sin 2x) + 9 sin x = -3multiply factors and group to obtain5 sin2x + 9 sin x -2 = 0let u = sinx and substitute to obtain anquadratic equation.5 u2 + 9 u - 2 = 0use any method to solve for u.By the quadratic formula, we obtain two solutions u1 and u2u1 = < -9 - sqrt(121) > / 10 = 1 / 5 = -2 and u2 = < -9 + sqrt(121) > / 10 = 0.2we now solve the equation for xu1 = sin x = -2
the above equation has no solutions for x since -2 is not in the range of values of sin(x); -1 u2 = sin x = 0.2sin x is positive in the first and secondquadrants of the unit circle. There are twosolutions to the above equation in <0 , 2 pi)(see unit circle below)
solution in the first quadrantx1 = arcsin 0.2 arcsin 0.2 is the inverse sine functionand it can be approximated using a calculator)solution in the second quadrantx2 = pi - arcsin 0.2conclusion: There are two solutions to the given equation.
Example 3Find all solutions for the trigonometric equationcot x cos 2x = cot x
Solution to example 3subtract cot x from both sides of the equation and simplifycot x cos 2x - cot x = 0Factor cot x cot x (cos 2x - 1) = 0Setting each factor in the above trigonometric equation to zero, we obtain two equations.cot x = 0 and cos 2x - 1 = 0The solutions to equation cot x = 0 are given byx = pi / 2 + k*pi , k is am integer. Equation cos 2x - 1 = 0 givescos x = 1 and cos x = -1The solutions to the above equations are given byx = 2k*pi and x = (2k + 1)*pi where k is an integerHOWEVER 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.
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Note that many of the techniques used in solvingalgebraic equations are also used to solve trigonometricequations.