
Find all solutions of the equation
sin(x) = 1 / 2
in the interval [0 , 2π) 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 = π / 6. (green in unit circle)

In quadrant II, and by symmetry of the unit circle, the solution to sin(x) = 1 / 2 is x = π  π / 6 = 5π/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:π / 6 and 5π/6.

Find all solutions of the equation
sin(x) + cos(x) = 0
in the interval [0 , 2π) 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 = 3π/4

By symmetry of the unit circle, the solution in quadrant IV is given by: 3π/4 + π = 7π / 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 3π/4 and 7π/4 are close to the x intercepts of the graph shown below.

Find all solutions of the equation:
sin(πx / 2)cos(π/3)  cos(πx / 2)sin(π/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(πx / 2)cos(π/3)  cos(πx / 2)sin(π/3) = sin(πx / 2  π/3)
 rewrite the given equation as follows: sin(πx / 2  π/3) = 0
 Solve the above equation to obtain: πx / 2  π/3 = k π where k = 0 , ~+mn~ 1 , ~+mn~ 2, ...

Simplify and rewrite the solution as: x = 2 k + 2/3 where k = 0 , ~+mn~ 1 , ~+mn~ 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(πx / 2)cos(π/3)  cos(πx / 2)sin(π/3) shown above.

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 π where k = 0 , ~+mn~ 1 , ~+mn~ 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} = 2π/3 + 2 k π where k = 0 , ~+mn~ 1 , ~+mn~ 2, ...

Third group of solutions: x_{3} = 4π/3 + 2 k π where k = 0 , ~+mn~ 1 , ~+mn~ 2, ...

The first 3 positive solutions are obtained by setting k = 0 in all three groups of solutions to obtain: 0, 2π/3 and 4π/3. These values are close to the x intercepts of the graph of g(x) = 2 sin^{2}(x) + cos (x)  1 shown above.

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) = ~+mn~ √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 = π / 6 + 2 k π where k = 0 , ~+mn~ 1 , ~+mn~ 2, ... (quadrant I)
and
x / 2 = 11π / 6 + 2 k π where k = 0 , ~+mn~ 1 , ~+mn~ 2, ... (quadrant VI)
Multiply all terms by 2 to obtain the first two groups of solutions:
x_{1} = π / 3 + 4 k π where k = 0 , ~+mn~ 1 , ~+mn~ 2, ...
x_{2} = 11π / 3 + 4 k π where k = 0 , ~+mn~ 1 , ~+mn~ 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 π / 6 + 2 k π where k = 0 , ~+mn~ 1 , ~+mn~ 2, ... (quadrant II)
and
x / 2 = 7 π / 6 + 2 k π where k = 0 , ~+mn~ 1 , ~+mn~ 2, ... (quadrant III)
Multiply all terms by 2 to obtain two more groups two of solutions:
x_{3} = 5 π / 3 + 4 k π where k = 0 , ~+mn~ 1 , ~+mn~ 2, ...
x_{4} = 7 π / 3 + 4 k π where k = 0 , ~+mn~ 1 , ~+mn~ 2, ...

The first 4 positive solutions are obtained by setting k = 0 in all four groups of solutions to obtain: π/3 , 5π/3, 7π/3 and 11π/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.

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} = 2π/3 + 2 k π where k = 0 , ~+mn~ 1 , ~+mn~ 2, ...
x_{2} = 4π/3 + 2 k π; where k = 0 , ~+mn~ 1 , ~+mn~ 2, ...

The first 2 positive solutions are obtained by setting k = 0 in the two groups of solutions to obtain: 2π/3 and 4π/3. These values are close to the x intercepts of the graph of f(x) = tan^{2} ( x) + 2 sec (x) + 1 shown above.

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} = kπ

cos(x) = 0 gives solutions: x_{2} = π/2 + kπ

1  2 sin(x) = 0 or sin(x) = 1/2 gives 2 other gropus of solutions.
x_{3} = π/6 + 2kπ
and
x_{4} = 5π/6 + 2kπ

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 , π, π/2, 3π/2, and k = 0 in the second and third groups of solutions x_{3} and x_{4} to obtain π/6 and 5π/6. These 6 solutions are close to the x intercepts of the graph of f(x) = cos(3x) + sin(2x)  cos(x) shown above.

Solve the equation: cos(3x) = cos(2x + π/4).
Solution
 We use the fact that if cos(A) = cos(B), then A = B + 2kπ or A =  B + 2kπ to write:
3 x = 2 x + π/4 + 2kπ
or
3 x = (2 x + π/4) + 2kπ

Solve 3 x = 2 x + π/4 + 2kπ Solutions: x_{1} = π/4 + 2kπ
 Solve 3 x = (2 x + π/4) + 2kπ ,
gives 5x =  π/4 + 2kπ

divide all terms by 5 : x_{2} =  π/20 + 2kπ / 5
where k = 0 , ~+mn~ 1 , ~+mn~ 2, ... in both groups of solutions

The 6 x intercepts shown in the graph of f(x) = cos(3x)  cos(2x + π/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} =  π/20 + 2kπ / 5.

