Solve Trigonometric Equations Questions with Detailed Solutions
Ways of solving trigonometric equations with detailed solutions are presented along with graphical interpretation.
Question 1
Find all solutions of the equation
sin(x) = 1 / 2
in the interval [0 , 2?) and explain the solutions graphically using a unit circle and the graph of sin (x) - 1/2 in a rectangular coordinate system of axes.
Solution
sin(x) is positive in quadrant I and II and therefore the given equation has two solutions.
In quadrant I, the solution to sin(x) = 1 / 2 is x = ? / 6. (green in unit circle)
In quadrant II, and by symmetry of the unit circle, the solution to sin(x) = 1 / 2 is x = ? - ? / 6 = 5?/6 (brown in unit circle)
The graphical solutions of the equation sin(x) = 1 / 2 are obtained by first writing the equation with right side equal to zero sin(x) - 1 / 2 = 0. Then graph the left side of the equation and the solutions of the equation are the x - intercepts of the graph of f(x) = sin(x) - 1 / 2 as shown below. It is easy to check that the two x intercepts in the graph below are close to the the analytical solutions found above:? / 6 and 5?/6.
Question 2
Find all solutions of the equation
sin(x) + cos(x) = 0
in the interval [0 , 2?) and explain the solutions graphically using the unit circle and the graph of sin (x) + cos(x) in a rectangular coordinate system.
Solution
The equation may be written as: sin(x) = -cos(x).
Divide both sides of the equation by cos(x): sin(x)/ cos(x) = -1
Use the identity tan(x) = sin(x)/cos(x) to rewrite the equation as: tan(x) = - 1
Solve the equation tan(x) = - 1
tan(x) is negative in quadrant II and IV
Solution in quadrant II: x = 3?/4
By symmetry of the unit circle, the solution in quadrant IV is given by: 3?/4 + ? = 7? / 4
The graphical solutions may be approximated by the x intercepts of the graph of g(x) = sin(x) + cos(x). We can easily check that the two solutions found above 3?/4 and 7?/4 are close to the x intercepts of the graph shown below.
Question 3
Find all solutions of the equation:
sin(?x / 2)cos(?/3) - cos(?x / 2)sin(?/3) = 0
and check graphically the first few positive solutions.
Solution
We first use the formula sin(A - B) = sin A cos B - cos A sin B
to write : sin(?x / 2)cos(?/3) - cos(?x / 2)sin(?/3) = sin(?x / 2 - ?/3)
rewrite the given equation as follows: sin(?x / 2 - ?/3) = 0
Solve the above equation to obtain: ?x / 2 - ?/3 = k ? where k = 0 , ± 1 , ± 2, ...
Simplify and rewrite the solution as: x = 2 k + 2/3 where k = 0 , ± 1 , ± 2, ...
The first 3 positive solutions are: 2/3 , 8/3 and 14/3 corresponding to k =0, 1 and 2. These values are close to the x intercepts of the graph of f(x) = sin(?x / 2)cos(?/3) - cos(?x / 2)sin(?/3) shown above.
Question 4
Find all solutions of the equations: 2 sin^{2}(x) + cos (x) = 1.
and check graphically the first few positive solutions.
Solution
We first use the identity: sin^{2}(x) = 1 - cos^{2}(x)
to rewrite the equation as: 2(1 - cos^{2}(x)) + cos (x) = 1
group like terms and write equation with right side equal to zero:
-2 cos^{2}(x) + cos (x) +1 = 0
Let u = cos (x) and rewrite the equation as: -2 u^{2} + u +1 = 0
Solve for u: u = 1 and u = - 1 / 2
First group of solutions: Solve u = 1 for x: u = cos (x) = 1 : x_{1} = 2 k ? where k = 0 , ± 1 , ± 2, ...
Solve u = -1 / 2 for x:
cos(x) is negative in quadrant II and III, hence u = cos (x) = - 1/2 has two more groups of solutions given by
Second group of solutions: x_{2} = 2?/3 + 2 k ? where k = 0 , ± 1 , ± 2, ...
Third group of solutions: x_{3} = 4?/3 + 2 k ? where k = 0 , ± 1 , ± 2, ...
The first 3 positive solutions are obtained by setting k = 0 in all three groups of solutions to obtain: 0, 2?/3 and 4?/3. These values are close to the x intercepts of the graph of g(x) = 2 sin^{2}(x) + cos (x) - 1 shown above.
Question 5
Find all solutions of the equation: 6 cos^{2}(x / 2) - cos (x) = 4.
and check graphically the first few positive solutions.
Solution
We first use the identity: cos(x) = 2 cos^{2}(x / 2) - 1
to rewrite the equation as: 6 cos^{2}(x / 2) - (2 cos^{2}(x / 2) - 1 ) = 4
group like terms and write equation with right side equal to zero:
4 cos^{2}(x / 2) = 3
cos(x / 2) = ± √3 / 2
Solve the first equation cos(x / 2) = √3 / 2
the cosine function is positive in quadrant I and IV hence the two groups of solutions:
x / 2 = ? / 6 + 2 k ? where k = 0 , ± 1 , ± 2, ... (quadrant I)
and
x / 2 = 11? / 6 + 2 k ? where k = 0 , ± 1 , ± 2, ... (quadrant VI)
Multiply all terms by 2 to obtain the first two groups of solutions:
x_{1} = ? / 3 + 4 k ? where k = 0 , ± 1 , ± 2, ...
x_{2} = 11? / 3 + 4 k ? where k = 0 , ± 1 , ± 2, ...
Solve the second equation cos(x / 2) = - √3 / 2
The cosine function is negative in quadrant II and III hence the two groups of solutions:
x / 2 = 5 ? / 6 + 2 k ? where k = 0 , ± 1 , ± 2, ... (quadrant II)
and
x / 2 = 7 ? / 6 + 2 k ? where k = 0 , ± 1 , ± 2, ... (quadrant III)
Multiply all terms by 2 to obtain two more groups two of solutions:
x_{3} = 5 ? / 3 + 4 k ? where k = 0 , ± 1 , ± 2, ...
x_{4} = 7 ? / 3 + 4 k ? where k = 0 , ± 1 , ± 2, ...
The first 4 positive solutions are obtained by setting k = 0 in all four groups of solutions to obtain: ?/3 , 5?/3, 7?/3 and 11?/3. These values are close to the x intercepts of the graph of f(x) = 6 cos^{2}(x / 2) - cos (x) - 4 shown above.
Question 6
Solve the equation: tan^{2} ( x) + 2 sec (x) + 1 = 0.
and check graphically the first few positive solutions.
Solution
We first use the identity: tan^{2}(x) = sec^{2}(x) - 1
to rewrite the equation as: sec^{2}(x) - 1 + 2 sec (x) + 1 = 0
group like terms and write equation with right side equal to zero:
sec^{2}(x) + 2 sec (x) = 0
Factor: sec (x) (sec (x) + 2) = 0
sec (x) = 0 has no solution.
Solve : sec (x) + 2 = 0
Use sec (x) = 1 / cos (x) to rewrite equation as : cos(x) = - 1 / 2
The cosine function is negative in quadrant II and III hence two groups of solutions
x_{1} = 2?/3 + 2 k ? where k = 0 , ± 1 , ± 2, ...
x_{2} = 4?/3 + 2 k ?; where k = 0 , ± 1 , ± 2, ...
The first 2 positive solutions are obtained by setting k = 0 in the two groups of solutions to obtain: 2?/3 and 4?/3. These values are close to the x intercepts of the graph of f(x) = tan^{2} ( x) + 2 sec (x) + 1 shown above.
Question 7
Solve the equation: cos(3x) + sin(2x) = cos(x).
Solution
We first use the identities: cos(3x) = cos^{3}(x) - 3 cos(x)sin^{2}(x)
and
sin(2x) = 2sin(x) cos(x)
to rewrite the given equation as follows: cos^{3}(x) - 3 cos(x)sin^{2}(x) + 2sin(x) cos(x) = cos(x)
Rewrite with right hand side equal to zero and factor cos(x)
cos(x) ( cos^{2}(x) - 3 sin^{2}(x) + 2 sin(x) - 1) = 0
Use the identity cos^{2}(x) = 1 - sin^{2}(x) to rewrite equation as follows
cos(x) (1 - sin^{2}(x) - 3 sin^{2}(x) + 2 sin(x) - 1) = 0
Simplify
cos(x) (- 4 sin^{2}(x) + 2 sin(x) ) = 0
Factor
2 sin(x) cos(x) (1 - 2 sin(x)) = 0
Set each factor equal to zero and solve.
sin(x) = 0 gives solutions: x_{1} = k?
cos(x) = 0 gives solutions: x_{2} = ?/2 + k?
1 - 2 sin(x) = 0 or sin(x) = 1/2 gives 2 other gropus of solutions.
x_{3} = ?/6 + 2k?
and
x_{4} = 5?/6 + 2k?
The first 6 positive solutions are obtained by setting k = 0 and k = 1 in the first two groups of solutions x_{1} and x_{2} to obtain: 0 , ?, ?/2, 3?/2, and k = 0 in the second and third groups of solutions x_{3} and x_{4} to obtain ?/6 and 5?/6. These 6 solutions are close to the x intercepts of the graph of f(x) = cos(3x) + sin(2x) - cos(x) shown above.
Question 8
Solve the equation: cos(3x) = cos(2x + ?/4).
Solution
We use the fact that if cos(A) = cos(B), then A = B + 2k? or A = - B + 2k? to write:
3 x = 2 x + ?/4 + 2k?
or
3 x = -(2 x + ?/4) + 2k?
Solve 3 x = 2 x + ?/4 + 2k? Solutions: x_{1} = ?/4 + 2k?
Solve 3 x = -(2 x + ?/4) + 2k? ,
gives 5x = - ?/4 + 2k?
divide all terms by 5 : x_{2} = - ?/20 + 2k? / 5
where k = 0 , ± 1 , ± 2, ... in both groups of solutions
The 6 x intercepts shown in the graph of f(x) = cos(3x) - cos(2x + ?/4) below correspond the solutions k = 0 in the first group x_{1} and k = 0, 1, 2 ,3 and 4 in the second group of solutions x_{2} = - ?/20 + 2k? / 5.