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Solve by Isolating Trigonometric Expressions
Solve .
Original equation
Subtract 3cos x from each side to isolate the trigonometric expression.
Solve for cos x.
Solve by Isolating Trigonometric Expressions
Answer:
The period of cosine is 2π, so you only need to find
solutions on the interval [0, 2π). The solutions on this
interval are . The solutions on the interval
(–∞, ∞) are then found by adding integer multiples 2π.
Therefore, the general form of the solutions is
x = , where n is an integer.
Solve sin x + = – sin x.
A.
B.
C.
D.
Solve by Taking the Square Root of Each Side
Solve 3 tan2 x – 4 = –3.
3 tan2 x – 4 = –3 Original equation
3 tan2 x = 1 Add 4 to each side.
Divide each side by 3.
Take the square root of each side.
Rationalize the denominator.
Solve by Taking the Square Root of Each Side
The period of tangent is π. On the interval [0, π),
tan x = when x = and tan x = when x = .
The solutions on the interval (–∞, ∞) have the general
form , where n is an integer.
Answer:
Solve 5 tan2x – 15 = 0.
A.
B.
C.
D.
Solve by Factoring
A. Find all solutions of on the interval [0, 2π).
Original equation
Isolate the trigonometric terms.
Factor.
Zero Product Property
Solve by Factoring
Solve for x on [0, 2π).
On the interval [0, 2π), the equation
has solutions .
Answer:
Solve for cos x.
Solve by Factoring
B. Find all solutions of 2sin2x + sinx – 1 = 0 on the interval [0, 2π).
Original equation
Factor.
Zero Product PropertySolve for sin x.
Solve for x on [0, 2π).
Solve by Factoring
Answer:
On the interval [0, 2π), the equation 2sin2x + sinx – 1 = 0
has solutions .
Find all solutions of 2 tan4 x – tan2 x – 15 = 0 on the interval [0, π).
A.
B.
C.
D.
Trigonometric Functions of Multiple Angles
PROJECTILES A projectile is sent off with an initial
speed vo of 350 m/s and clears a fence 3000 m
away. The height of the fence is the same height as
the initial height of the projectile. If the distance the
projectile traveled is given by , find the
interval of possible launch angles to clear the
fence.
Trigonometric Functions of Multiple Angles
Original formula
d = 3000 and v0 = 350
Simplify.
Multiply each side by 9.8.
Divide each side by 122,500.
Definition of inverse sine.
Trigonometric Functions of Multiple Angles
Recall from Lesson 4-6 that the range of the inverse sine function is restricted to acute angles of θ in the interval [–90°, 90°]. Since we are finding the inverse sine of 2θ instead of θ, we need to consider angles in the interval [–2(90°), 2(90°)] or [–180°, 180°]. Use your calculator to find the acute angle and the reference angle relationship sin (180° − θ) = sin θ to find the obtuse angle.
sin–10.24 = 2 Definition of inverse sine
13.9° or 166.1°= 2sin–1(0.24) ≈13.9° and sin(180° – 13.9°) = 166.1°
7.0° or 83.1° = Divide by 2.
Answer: 7.0° ≤ ≤ 83.1°
Trigonometric Functions of Multiple Angles
The interval is [7.0°, 83.1°]. The ball will clear the fence if the angle is between 7.0° and 83.1°.
CHECK Substitute the angle measures into the original equation to confirm the solution.
Original formula
Use a calculator.
= 7.0° or = 83.1°
GOLF A golf ball is sent off with an initial speed vo
of 36 m/s and clears a small barricade 70 m away.
The height of the barricade is the same height as
the initial height of the ball. If the distance the ball
traveled is given by , find the interval
of possible launch angles to clear the barricade.
A. 1.6° ≤ ≤ 88.5°
B. 3.1° ≤ ≤ 176.9°
C. 16.0° ≤ ≤ 74.0°
D. 32° ≤ ≤ 148.0°
Solve by Rewriting Using a Single Trigonometric Function
Find all solutions of sin2 x – sin x + 1 = cos2 x on the interval [0, 2π).
sin2 x – sin x + 1
= cos2 x
Original equation
–cos2 x + sin2 x – sin x + 1
= 0
Subtract cos2 x from each side.
–(1 – sin2 x) + sin2 x – sin x + 1
= 0
Pythagorean Identity
2sin2 x – sin x
= 0
Simplify.
sin x (2sin x – 1)
= 0
Factor.
Solve by Rewriting Using a Single Trigonometric Function
sin x = 02sin x – 1= 0Zero Product Property
2sin x = 1Solve for sin x.
Solve for x on [0, 2π).
x = 0, π
Solve by Rewriting Using a Single Trigonometric Function
CHECK The graphs of Y1 = sin2 x – sin x + 1 and
Y2 = cos2 x intersect at on the interval
[0, 2π) as shown.
Answer:
Find all solutions of 2sin2x = cosx + 1 on the interval [0, 2).
A.
B.
C.
D.
Find all solutions of sin x – cos x = 1 on the interval [0, 2π).
sin x – cos x= 1Original equation
sin x= cos x + 1Add cos x to each side.
sin2 x= cos2 x + 2cos x + 1 Square each side.
1 – cos2 x= cos2 x + 2cos x + 1Pythagorean Identity
0= 2cos2 x + 2cos xSubtract 1 – cos2x from each side.
0= cos2 x + cos x Divide each side by 2.
0= cos x(cos x + 1)Factor.
Solve by Squaring
Solve by Squaring
cos x = 0 cos x + 1= 0 Zero Product Property
cos x= –1 Solve for cos x.
Original formula
Simplify.
, x = πSolve for x on [0, 2).
Substitute sin π – cos π = 1
Solve by Squaring
Therefore, the only valid solutions are on the interval .
Answer:
Find all solutions of 1 + cos x = sin x on the interval [0, 2π).
A.
B.
C.
D.
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