10.4 Solve Trigonometric Equations
Solving trig equations…
What’s the difference?
Solve a trig equation
Solve 2 sin x – 3 = 0.
SOLUTION
First isolate sin x on one side of the equation.
2 sin x – 3 = 0
2 sin x = 3
sin x = 3
2
One solution of sin x =
Write original equation.
Add 3 to each side.
Divide each side by 2.
3
2 in the interval 0 ≤ x < 2 π is
π.
x = sin–1 3 =
3
2
The other solution in the interval is x = π – π = 2π .
3
3
Moreover, because y = sin x is periodic, there will be
infinitely many solutions.
You can use the two solutions found above to write the
general solution:
π
2π
x = + 2nπ
x = + 2nπ (where n is any integer)
or
3
3
CHECK
Solve a trigonometric equation in an interval
Solve 9 tan2 x + 2 = 3 in the interval 0 ≤ x <2π.
9 tan2 x + 2 = 3
Write original equation.
9 tan2 x = 1
Subtract 2 from each side.
1
2
tan x = 9
Divide each side by 9.
Take square roots of each side.
tan x = +– 1
3
Using a calculator, you find that
tan –1 1 0.322 and tan –1 (– 1 ) – 0.322.
3
3
Therefore, the general solution of the equation is:
x 0.322 + nπ or x – 0.322 + nπ (where n is any integer)
ANSWER
The specific solutions in the interval 0 ≤ x <2π are:
x 0.322
x – 0.322 + π
2.820
x 0.322 + π 3.464
x – 0.322 + 2π 5.961
1. Find the general solution of the equation
2 sin x + 4 = 5.
2 sin x + 4 = 5
Write original equation.
2 sin x = 1
sin x = 1
2
Subtract 4 from each side.
, 5π
sin-1 x = π
3 6
Use a calculator to find both
solutions within 0 < x < 2π
ANSWER
π + 2n π or 5π + 2n π
3
6
Divide both sides by 2.
2. Solve the equation 3 csc2 x = 4 in the interval 0 ≤ x <2π.
3 csc2 x = 4
4
csc2 x =
3
csc x = 2
√3
1
2
=
sin x √3
√3
sin x =
2
√3
-1
sin x =
2
ANSWER
Write original equation.
Divide both sides by 3
Take the square root of both sides
Reciprocal identity
Standard Form
Use a calculator to find all
solutions within 0 < x < 2π
π , 2π, 4π, 5π
3 3 3 3
Oceanography
The water depth d for the Bay of Fundy can be
modeled by
d = 35 – 28 cos π t
6.2
where d is measured in feet and t is the time in
hours. If t = 0 represents midnight, at what time(s) is
the water depth 7 feet?
SOLUTION
Substitute 7 for d in the model and solve for t.
35 – 28 cos π t = 7
Substitute 7 for d.
6.2
– 28 cos π t = –28
Subtract 35 from each side.
6.2
cos π t = 1
Divide each side by –28.
6.2
π t = 2nπ
cos q = 1 when q = 2nπ.
6.2
t = 12.4n
Solve for t.
ANSWER
On the interval 0≤ t ≤ 24 (representing one full day), the
water depth is 7 feet when t = 12.4(0) = 0 (that is, at
midnight) and when t = 12.4(1) = 12.4 (that is, at 12:24 P.M.).
10.4 Assignment
Page 639, 3-21 all
10.4 Solve Trigonometric Equations
sin3 x – 4 sin x = 0
sin x (sin2 x – 4) = 0
sin x (sin x + 2)(sin x – 2) = 0
SOLUTION
Write original equation.
Factor out sin x.
Factor difference of squares.
Set each factor equal to 0 and solve for x, if possible.
sin x + 2 = 0
sin x – 2 = 0
sin x = 0
x = 0 or x = π
sin x = –2
sin x = 2
The only solutions in the interval 0 ≤ x ≤ 2π, are x = 0 and x = π.
The general solution is x = 2nπ or x = π + 2nπ where n is any integer.
ANSWER
The correct answer is D.
Eliminate solutions
Because sin x is never less than −1 or
greater than 1, there are no solutions of
sin x = −2 and sin x = 2.
The same is true with the cosine of x is
never greater than 1.
Use the quadratic formula
Solve cos2 x – 5 cos x + 2 = 0 in the interval 0 ≤ x <π.
2
SOLUTION Because the equation is in the form au + bu + c = 0
where u = cos x, you can use the quadratic formula
to solve for cos x.
cos2 x – 5 cos x + 2 = 0
Write original equation.
cos x =
=
– (–5) +– (–5)2 – 4(1)(2)
2(1)
5 +– 17
2
4.56 or 0.44
No solution
ANSWER
x = cos –1 0.44
1.12
Quadratic formula
Simplify.
Use a calculator.
Use inverse cosine.
Use a calculator, if possible
In the interval 0 ≤ x ≤ π, the only solution is x
1.12.
Solve an equation with an extraneous solution
Solve 1 + cos x = sin x in the interval 0 ≤ x < 2π.
1 + cos x = sin x
(1 + cos x)2 = (sin x)2
Write original equation.
Square both sides.
1 + 2 cos x + cos2 x = sin2 x
Multiply.
1 + 2 cos x + cos2 x = 1– cos2 x
Pythagorean identity
2 cos2 x + 2 cos x = 0
Quadratic form
2 cos x (cos x + 1) = 0
Factor out 2 cos x.
2 cos x = 0 or cos x + 1 = 0
Zero product property
cos x = 0 or cos x = –1
Solve for cos x.
On the interval 0 ≤ x <2π, cos x = 0 has two solutions:
π
3π
x = 2 or x = 2
On the interval 0 ≤ x <2π, cos x = –1 has one solution: x = π.
Therefore, 1 + cos x = sin x has three possible solutions:
π
3π
x = 2 , π, and 2
CHECK To check the solutions, substitute them into
the original equation and simplify.
1 + cos x = sin x
1 + cos x = sin x
1 + cos x = sin x
?
3π ?
3π
π
π ?
1
+
cos
π
1 + cos
1 + cos 2 = sin 2
= sin π
2 = sin 2
?
?
?
1+0 =1
1 + (–1) = 0
1 + 0 = –1
0 =0
1 = 1
1 = –1
ANSWER
ANSWER
Graphs of each side of the original equation confirm
the solutions.
Find the general solution of the equation.
4. sin3 x – sin x = 0
sin3 x – sin x = 0
Write original equation.
sin x(sin2 x – 1) = 0
Factor
sin x(– cos2 x ) = 0
Pythagorean Identity
sin x = 0 OR (– cos2 x ) = 0
Zero product property
sin x = 0 OR cos x = 0
Solve
On the interval 0 ≤ x <2π, sin x = 0 has two solutions:
x = 0, π.
On the interval 0 ≤ x <2π, cos x = 0 has two solutions:
x = π , 3π
2 2
0 + n π or π + n π
ANSWER
2
Find the general solution of the equation.
5. 1 – cos x = 3 sin x
1 – cos x = 3 sin x
Write original equation.
1 – 2 cos x + cos2 x = 3 sin2 x
Square both sides.
1 – 2 cos x + cos2 x = 3 (1 – cos2 x)
Pythagorean Identity
1 – 2 cos x + cos2 x = 3 – 3 cos2 x
– 2 – 2 cos x + 4 cos2 x = 0
– 2(1 + cos x – 2 cos2 x) = 0
– 2(1 + 2 cos x) (1 – cos x) = 0
(1 + 2 cos x) = 0 OR (1 – cos x) = 0
1
cos x = – 2 OR cos x = 1
Distributive Property
Group like terms
Factor
Factor
Zero product property
Solve for cos x
1
On the interval 0 ≤ x < 2π, cos x = – 2has two solutions
x = 2π 4π
3 3
On the interval 0 ≤ x < 2π, cos x = 1 has one solution:
x = 0.
4π
But sine is negative in Quadrant III, so 3 is not a solution
ANSWER
0 + 2n π or 2π+ 2n π
3
Solve the equation in the interval 0 ≤ x < π.
6. 2 sin x = csc x
2 sin x = csc x
Write original equation.
2 sin x = sin1 x
Reciprocal identity
2 sin2 x = 1
sin2 x = 1
2
sin x = √2
2
Multiply both sides by sin x
Divide both sides by 2
Take the square root of both sides
π , 3π
4 4
ANSWER
π , 3π
4 4
Solve the equation in the interval 0 ≤ x <π.
7. tan2 x – sin x tan2 x = 0
tan2 x – sin x tan2 x = 0
tan2 x (1 – sin x) = 0
tan2 x = 0 OR (1 – sin x) = 0
tan x = 0 OR sin x = 1
Write original equation.
Factor
Zero product property
Simplify
On the interval 0 ≤ x < π, tan x = 0 has two solutions x = 0, π
ANSWER 0, π or π
2
10-4 Assignment day 2
Page 639, 24-35 all