Solving Trigonometric Equations

```Solving Trigonometric
Equations
 Trig identities are true for all values of the
variable for which the variable is defined.
 However, trig equations, like algebraic
equations, are true for some but not all values
of the variable.
 Trig equations do not have unique solutions.
 Trig equations have infinitely many solutions.
 They differ by the period of the function, 2p or
360&deg; for sin and cos, and p or 180&deg; for tan.
Solving Trig Equations
 Trig equations will have unique solutions if the
value of the function is restricted to two
 These solutions are called the principal values.
 For sin x and tan x, the principal values are in
Quadrants I or IV. x is in the interval
-90&deg; &lt; x &lt; 90&deg;.
 For cos x, the principal values are in Q I or II.
 So x is in the interval 0&deg; &lt; x &lt; 180&deg;.
Solve 2 cos&sup2; x – 5 cos x + 2 = 0
for the principal values of x.
2 cos&sup2; x – 5 cos x + 2 = 0
(2 cos x - 1) (cos x – 2) = 0
2 cos x - 1 = 0
or
Factor
cos x – 2 = 0
2 cos x = 1
cos x = 2
cos x = &frac12;
There is no solution for cos x = 2
since –1 &lt; cos x &lt; 1
x = 60&deg;
Solve 2 tan x sin x + 2 sin x = tan x + 1
for all values of x.
2 tan x sin x + 2 sin x = tan x + 1
Subtract tan x + 1 from
2 tan x sin x + 2 sin x – tan x – 1 = 0 both sides.
(tan x + 1) (2 sin x – 1) = 0
factor
When all the values
tan x + 1 = 0
or
2 sin x – 1 = 0 of x are required,
the solution should
2 sin x = 1
tan x = -1
be represented as
sin x = &frac12;
x + 360k&deg; for sin
x = -45&deg; + 180k&deg;
x = 30&deg; + 360k&deg; and cos, and
x + 180k&deg; for tan,
or
x = 150&deg; + 360k&deg; where k is any
integer.
Solve sin&sup2; x + cos 2x – cos x = 0
for the principal values of x.
sin&sup2; x + cos 2x – cos x = 0
sin&sup2; x + (1 – 2 sin&sup2; x) – cos x = 0
Express cos 2x in terms
of sin x.
1 – sin&sup2; x – cos x = 0 Combine like terms
1 – sin&sup2; x = cos&sup2; x
cos&sup2; x – cos x = 0
Factor
cos x (cos x – 1) = 0
So the solutions
or
cos x = 0
cos x – 1 = 0 are 0&deg; and 90&deg;
cos x = 1
x = 90&deg;
x = 0&deg;
Solve cos x = 1 + sin x
for 0&deg; &lt; x &lt; 360&deg;
cos x = 1 + sin x
cos&sup2; x = (1 + sin x)&sup2; Square both sides
cos&sup2; x = 1 + 2 sin x + sin&sup2; x Expand the binomial squared
1 – sin&sup2; x = 1 + 2 sin x + sin&sup2; x cos&sup2; x = 1 – sin&sup2; x
0 = 2 sin x + 2 sin&sup2; x
0 = 2 sin x (1 + sin x)
Factor
or
2 sin x = 0
1 + sin x = 0
sin x = -1
sin x = 0
x = 270&ordm;
x = 0&deg; or 180&ordm;
It’s important to always check your
solutions. Some may not actually be
solutions to the original equation.
x = 0&ordm; or 180&ordm;
x = 270&ordm;
cos x = 1 + sin x
cos 0&ordm; = 1 + sin 0&deg;
cos 180&ordm; = 1 + sin 180&ordm;
1=1 ☻
-1 = 1 + 0
-1 ≠ 1
cos 270&ordm; = 1 + sin 270&ordm; Based on the check,
0 = 1 + (-1)
180&ordm; is not a solution
0=0 ☻
Assignment
Page 390
– # 17 - 38
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