3.7 SOLVING TRIG EQUATIONS
Recall periods of sin and cos:
cos (θ) = cos(θ ± 2πk) and sin (θ) = sin (θ ± 2πk)
for any integer, k.
Example 2 on p. 269
How do you solve cos θ = ½ for θ?
You can take the arccos of both sides to get θ by itself.
cos-1(cos θ) = θ = cos-1( ½) = π/3
However, arccos only gives us an answer between 0 and π.
There are other angles that give us a cosine value of ½ .
Look at the unit circle diagram. What other angle gives a cosine
value of ½?
cos π/3 =
cos 5π/3 = ½
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We see from the unit circle that there are two possibilities for θ between
0 and 2π for which cos θ = ½:
θ = π/3 or θ = 5π/3
Since the values of cosine are repeated for integral multiples of 2π, other
values of θ that solve cos θ = ½ are:
θ = π/3 ± 2πk or θ = 5π/3 ± 2πk, where k is any integer.
NOW YOU DO #31 on p.272
Example 3:
Solve the equation 2 sin θ + 3 = 0, 0 ≤ θ < 2π
Solution: First let x = sin θ, and solve for x. Then solve for θ.
The equation becomes 2 x + 3 = 0
2x = x =
-
3
3
2
Now replace
sin θ =
-
x with
sin θ and solve for θ
3
2
Since sin θ < 0, we know θ can only be in Quadrant III or IV.
We choose 4π/3 and
5π/3 because they
meet the
requirement of
0 ≤ θ < 2π
sin 4π/3 =
- 3
2
sin 5π/3 =
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- 3
2
NOW YOU
DO #7
Example 4
sin (2θ) = ½ , 0 ≤ θ < 2π
Don’t convert this to the double-angle formula because we
don’t know what sin (θ) is.
Solve this problem the way it is. Find angle 2θ that has a sine
value of ½, then divide that angle by 2.
2θ = π/6 ± 2πk or 2θ = 5π/6 ± 2πk
Dividing both sides by 2 gives us:
π
1
π
 1 
θ =   2θ =  ± 2 π k    =
± πk
2
 6
  2  12
OR
1
 5π
  1  5π
θ =   2θ = 
± 2π k    =
± πk
2
 6
  2  12
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Example 5
Solve the equation:
π

tan θ −  = 1, 0 ≤ θ ≤ 2π
2

Now, tan has a period of π, so tan(θ) = tan(θ± πk) for any integer
k.
π
θ − = tan −1 (1) ± kπ
2
π π
θ − = ± kπ
2 4
π π
3π
θ = + ± kπ =
± kπ
4 2
4
Check k=0,1,2, etc.. until you reach a θ>2π [since our problem
limits 0 ≤θ<2π.
k=0: θ = 3π ± (0)π = 3π
4
4
k=1: θ = 3π ± (1)π = 7π
4
4
k=2: θ = 3π ± (2)π = 11π > 2π so this θ is not in given interval.
4
4
EXAMPLE 6 DONE IN CLASS
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HOMEWORK:
P. 272
#7-67 EOO (e.g. #7,11,15,19,23, etc..)
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