14.4 Solve trigonometric equations.

Your Notes

Goal · Solve trigonometric equations.

Example 1

Solve a trigonometric equation in an interval

Solve 2 cos² x + 1 = 2 in the interval 0 < x < 3tt.

Solution

2 cos^2 x + 1 = 2		Write original equation.
2 cos^2 x = _1_		Subtract 1 from each side.
cos^2x = ______		Divide each side by 2.
cos x = ______		Take square roots of each side.
x = cos ^-1 _____	or	x = cos^-1_____
x = or x = -		x = or x =

Therefore, the general solution of the equation is:

x = + _2np_	or	x = - + _2np_ or
x = + _2np_	or	x = + _2np_

To write the general solution of a trigonometric equation, you can add multiples of the period to all the solutions from one cycle.

where n is any integer.

The specific solutions that are in the interval 0 £ x £ 3p are:

x = _____	x = _____
x = + 2p = _____	x = + 2p = _____
x = - + 2p = _____	x = _____

Your Notes

Example 2

Solve a trigonometric equation in an interval

Solve = 20 - 12 sin in the interval 0 < t < 24 when d = 8.

Solution

20 - 12 sin = 8	Substitute 8 for d.
-12 sin = _-12_	Subtract 20 from each side.
sin = _1_	Divide each side by -12.
= + _2np_	sin q = 1 when q = + _2np_
t = _2 + 8n_	Solve for f.

On the interval 0 < t < 24, d is 8 when

t = _2 + 8(0)_ = _2_, t = _2 + 8(1)_ = _10_, and t = _2 + 8(2)_ = _18_.

Example 3

Use the quadratic formula

Solve 2 sin² x + 5 sin x + 3 = 0 in the interval -p < x < p.

Solution

2 sin^2 x + 5sinx + 3 = 0		Write original equation.
Sin x =		Quadratic formula
=		Simplify.
= _-1_ or _-1.5_		Simplify.
x = sin^-1 _(-1)_ or x = sin^-1 _(-1.5)_		Use inverse sine.
= _____	__ No solution__	Use a calculator.

In the interval -it <x < it, the only solution is x = _{_____}

Your Notes

Checkpoint Solve the trigonometric equation in the interval.

1. 16 sin² x + 5 = 6; 0 < x < 7p

about 0.253, about 2.889

2. 20 - 12 sin = 25; 0 < t < 3p

about 4.547, about 7.452

3. cos² x + 3 cos x - 4 = 0;0 < x < p

0

Your Notes

Example 4

Solve an equation with an extraneous solution

Solve 1 - cos x = sin x in the interval 0 < x < p.

Solution

1 - cos x = sin x

(1 - cos x)² = _( sin x)²_

1-2 cos x + cos² x = _3 sin² x_

1-2 cos x + cos² x = 3 _(1 - cos² x)_

1 - 2 cos x + cos²x = _3_ - _3 cos² x_

_4 cos^2x - 2 cos x - 2_ = 0		Quadratic form
_2 cos^2 x - cos x - 1_ = 0		Divide each side by 2.
_(2 cos x + 1)(cos x - 1)_ = 0		Factor.
_2 cos x +1= 0_ or _cos x - 1= 0_		Zero product property
cos x = _____	or cos x = _1_	Solve for cos x.
x = _____ or x = _____	x = _0_	Solve for x.

The apparent solution _{_______} does not check in the original equation. The only solutions in the interval 0 < x < 2p are x =_0_ and x = _{_____}.

Checkpoint Complete the following exercise.

4. Solve the equation in Example 4 in the interval 0 < x < 4p

0, , 2p,