3.7 SOLVING TRIG EQUATIONS
Example 2 on p. 248
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 =
y
x
cos 5π/3 = ½
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 and 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 #1 on p.251
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.
y
x
We choose 4π/3 and
5π/3 because they
meet the
requirement of
0 ≤ θ < 2π
sin 4π/3 =
- 3
2
sin 5π/3 =
- 3
2
NOW YOU
DO #11
Example 4
sin (2θ) = ½ , 0 ≤ θ < 2π
Don’t convert this to the double-angle formula.
Solve this the way it is. Find angle that has a sine value of ½, then
divid that angle by 2.
2θ = π/6 + 2πk or 2θ = 5π/6 + 2πk
Dividing both sides by 2 gives us:
⎛1⎞
⎛π
⎞⎛ 1 ⎞ π
θ = ⎜ ⎟ 2θ = ⎜ + 2 π k ⎟ ⎜ ⎟ =
+ πk
2
6
2
12
⎝ ⎠
⎝
⎠⎝ ⎠
OR
⎞ ⎛ 1 ⎞ 5π
⎛ 5π
⎛1⎞
+ πk
+ 2π k ⎟ ⎜ ⎟ =
⎟ 2θ = ⎜
2
6
2
12
⎠⎝ ⎠
⎝
⎝ ⎠
Because dividing by 2 gives us a repeating cycle of πk, there are
multiple possible answers within the range of 0 and 2π. We
cannot use k ≤-1, because that would give is a θ < 0, and we
cannot use k=2, because that would give us a θ > 2π. So let’s see
what we get with k = 0 and k=1:
θ =⎜
k = 0:
θ=
π
12
k = 1:
θ=
π
12
+ π (0) =
+ π (1) =
π
12
π
12
+0=
+
π
12
θ=
12π 13π
=
12
12
So the complete solution of sin(2 θ ) =
θ=
π 5π 13π 17π
,
,
,
12 12 12 12
5π
5π
5π
+ π (0) =
+0=
12
12
12
θ=
5π
5π 12π 17π
+ π (1) =
+
=
12
12
12
12
1
, 0 ≤ θ < 2π is :
2
NOW YOU DO #17
Example 5
π⎞
⎛
tan⎜θ − ⎟ = 1, 0 ≤ θ < 2π
2⎠
⎝
Let’s see what (θ - π/2) must be for the tangent function to be 1.We
know that tan π/4 = 1 and the tangent value repeats itself after a
period of π.
Solve the equation
π⎞
⎛
tan⎜θ − ⎟ = 1
2⎠
⎝
For tangent, the cycle repeats for πk, so
θ−
π
2
π
4
+
π
+ πk
3π
2π
+ πk =
+ πk
4
4 2
4 4
As before, k ≤ -1 would give us θ < 0 and k ≥ 2 would give us
θ=
π
=
+ πk =
π
+
θ > 2π . So let' s try k = 0, k = 1
k =0
3π
3π
θ=
+ π ( 0) =
4
4
k =1
3π
3π 4π 7π
+ π (1) =
+
=
4
4
4
4
so the complete solution for 0 ≤ θ < 2π is
θ=
θ=
3π 7π
,
4 4
Example 6
When using a calculator to find the solution for θ, be aware that using the
inverse trig function only gives you an answer within the range of the
inverse function.
Solve the equation
sin θ = 0.3, 0 ≤ θ < 2π
θ = sin-1 (0.3) will give you an answer with –π/2 ≤ θ ≤ π/2
θ ≈ 0.304692654 ≈ .30
But what about Quadrants II and III? Sine is only positive in Quadrants I
and II.
From the supplementary angle theorem,
sin(π – θ) = sin θ
So π – θ ≈ π – 0.30 ≈2.84
So the complete solution is θ ≈ .30, 2.84
#40
sin θ = - 0.2, 0 ≤ θ < 2π
θ = sin-1 (-0.2) will give you an answer with –π/2 ≤ θ < 0 since the sine is
negative.
θ = sin-1 (-0.2) ≈ -0.2013579208 ≈ -.20
What are the equivalent angles in the range 0 ≤ θ < 2π that will give
sin θ = -.2?
We know sine is only negative in Quadrants III and IV. We also know sine repeats
itself for cycles of 2πk. Since θ ≈ -.20, θ + 2π will still be < 2π.
π y
sin (π+.2rad) ≈ -.2
π+.2rad ≈ 3.34rad
θ ≈.2rad
x
sin (-.2rad) ≈ -.2
-.2rad+ 2π ≈ 6.08
Solution:
θ = 3.34rad, 6.08rad
Snell’s Law of Refraction (Applied Problem #44 on p.252)
angle of incidence
Incident ray’s speed
sin θ1 v1
= = index of refraction
sin θ 2 v2
angle of incidence
Refracted ray’s speed
44. The index of refraction of light passing from a
vacuum into dense glass is 1.66. If the angle of
incidence is 50, determine the angle of refraction.
θ1 = 50°,
v1
= 1.66
v2
sin 50°
= 1.66
sin θ 2
cross multiply to get
1.66 sin θ 2 = sin 50°
Divide both sides by 1.66 to get Make sure to change mode to Degrees in
your calculator before computing sin 50°
sin 50°
sin θ 2 =
≈ .4615
1.66
Using the inverse sine function on the calculator we get
θ 2 = sin −1 sin(θ 2 ) = sin −1 (.4615) ≈ 27.5°
HOMEWORK
p.251
#11,15,17,19,23,35,41,43
3.8 Solving Trig Equations (II)
Example 1
Solve 2sin2θ – 3sinθ + 1 = 0, 0 ≤ θ < 2π
How do we get sin θ by itself to solve for θ?
Let’s let x = sin θ
2x2 – 3x + 1 = 0
We can factor to solve for x
(2x -1)(x -1) = 0
The zero-product property that either factor must be 0 for the product to be
0.
Alternate Method: Use Quadratic formula
2x – 1 = 0
2x = 1
If you have an equation in the form ax2 + bx + c = 0
x=½
The solution is:
sin θ = ½
2 different real solutions if b2-4ac >0
θ = π/6, 5π/6
1 repeated real solution if b2 – 4ac =0
− b ± b 2 − 4ac
or
x=
No real solutions if b2-4ac <0
2a
x–1=0
x=1
2 x 2 − 3 x + 1 = 0 gives a = 2, b = -3, c = 1
sin θ = 1
− (−3) + (−3) 2 − 4(2)(1) 3 + 1 4
θ = π/2
x=
2(2)
=
4
=
4
=1
OR
− (−3) − (−3) 2 − 4(2)(1) 3 − 1 2 1
x=
=
= =
2(2)
4
4 2
Example 2
Solve the equation 3cosθ + 3 = 2sin2θ , 0 ≤ θ < 2π
This example has a combination of both sine and cosine
functions, so making x=sinθ will not give me everything in
terms of x. However, we can identities to get everything
in terms of the same trig function, then convert to x. Let’s
use the fact that sin2θ + cos2θ = 1 and get
sin2θ = 1- cos2θ
We can not get everything in terms of cos θ
3cosθ + 3 = 2(1- cos2θ)
3cosθ + 3 = 2- 2cos2θ
Putting everything on the left-hand-side and leaving righthand-side with 0 gives me:
2cos2θ +3cosθ +1 = 0
Now let’s make x = cosθ
2x2+3x + 1 = 0
Factoring we get
(2x + 1)(x + 1) = 0
2x + 1 = 0
x+1=0
2x = -1
x=-1
OR
x=-½
cos θ = - 1
cos θ = - ½
θ=π
θ = 2π/3, 4π/3
Example 5
Solve the equation sin θ cos θ = - ½ , 0 ≤ θ < 2π
There is no identity to convert sin θ into cos θ, or vice versa,
unless we used cos θ = ± 1− sin 2 θ
but that would make the problem more complicated, so instead
let’s use sin(2θ) = 2 sin θ cos θ
Notice that the term (sin θ cos θ ) is also in our equation to
solve. Let’s get it by itself in this identity by dividing both
sides by 2:
½ sin(2θ) = sin θ cos θ
Now let’s substitute that for sin θ cos θ in our equation to solve:
½ sin(2θ) = - ½ Multiplying both sides by 2 gives
sin(2θ) = -1
2θ = 3π/2 + 2πk Divide both sides by 2 to get θ by itself
θ = 3π/4 + πk
We cannot use k ≤-1, because that would give is a θ < 0, and
we cannot use k=2, because that would give us a θ > 2π. So
let’s see what we get with k = 0 and k=1:
k=0:
θ = 3π/4 + π(0) = 3π/4
k=1:
θ = 3π/4 + π(1) = 3π/4 + 4π/4 = 7π/4
HOMEWORK
p.258
#1,3,7,17,19,25,57