| Solving trigonometric equations is very similar to solving
other algebraic equations. The same techniques are used.
The goal is to isolate the trig function. Example:
2 sin x – 1 = 0
2 sin x = 1
sin x = 1/2
The equation sin x = 1/2 has the solutions x =  /6
and x = 5/6
between 0 and 2.
There is an infinite amount of solutions because sin
x has a period of 2:
x = /6
+ 2n and
x = 5/6
+ 2n
where n is any integer.
The same rules apply when dealing with square roots
in trigonometric functions:
4 sin² x – 3 = 0
4 sin² x = 3
sin² x = 3/4
sin x = ± v(3)/2
x = /3
+ 2n
x = 2/3
2n
Equations that have more than one trigonometric function
must be factored. All like terms must be collected on
one side. Some of the factors may have no solutions.
sin² = 2 sin x
sin²x – 2 sin x = 0
sin x (sin x – 2) = 0
To solve each factor, set them equal to zero:
| sin x = 0 |
sin x – 2 = 0 |
| x = 0 + 2n |
sin x = 2 |
| |
no solution |
Quadratic Trigonometric Equations
|
|
Trig equations can be quadratic, for example:
2 sin²x – sin x – 1 = 0
With quadratic trig equations, the same rules apply as
with regular quadratic equations. Solve them by factoring
or using the quadratic equation. Solve:
2 sin²x – 3 sin x + 1 = 0 in the interval
[0, 2]
(2 sin x – 1)(sin x – 1) = 0
| 2 sin x – 1 = 0 |
sin x – 1 = 0 |
| sin x = 1/2 |
sin x = 1 |
| x = /6,
5/6 |
x = /2 |
When factoring, make sure that each quadratic equation
has only one trig function i.e. sine, cosine, etc. If
it does not have only one function, use the basic trig
identities to change the equation.
3 sec² x – 2 tan²x – 4 = 0
3 (tan² x + 1) – 2 tan² x – 4 =
0
tan² x – 1 = 0
tan² x = =1
tan x = ± 1
x = /4 +
n
x = 3/4
+ n
Sometimes an equation cannot be solved
as is and must be squared to make it quadratic. Extra
solutions are likely to become apparent that may not
be applicable to the original equation.
sin x + 1 = cos x
sin²x + 2 sin x + 1 = cos²x
sin²x + 2 sin x + 1 = 1 – sin²x
2 sin²x + 2 sin x = 0
2 sin x (sin x + 1) = 0
| 2 sin x = 0 |
sin x + 1 = 0 |
| sin x = 0 |
sin x = – 1 |
| x = 0, |
x = 3/2 |
Because the original equation was squared, each solution
must be checked and confirmed with the original equation.
The value
does not fit for x because:
sin + 1
? cos
0 + 1 ? – 1
Therefore, the only solutions in the interval [0, 2),
the only solutions are x = 0, 3/2. |
