Transcript Trigonometric Equations I

6
Inverse
Circular
Functions
and
Trigonometric
Equations
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Inverse Circular Functions
6 and Trigonometric Equations
6.1 Inverse Circular Functions
6.2 Trigonometric Equations I
6.3 Trigonometric Equations II
6.4 Equations Involving Inverse
Trigonometric Functions
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6.2 Trigonometric Equations I
Solving by Linear Methods ▪ Solving by Factoring ▪ Solving by
Quadratic Methods ▪ Solving by Using Trigonometric Identities
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Solving a Trigonometric Equation
 Decide whether the equation is linear or
quadratic in form, so you can determine
the solution method.
 If only one trigonometric function is
present, first solve the equation for that
function.
 If more than one trigonometric function
is present, rearrange the equation so
that one side equals 0. Then try to factor
and set each factor equal to 0 to solve.
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Solving a Trigonometric Equation
 If the equation is quadratic in form, but
not factorable, use the quadratic
formula. Check that solutions are in the
desired interval.
 Try using identities to change the form of
the equation. If may be helpful to square
both sides of the equation first. If this is
done, check for extraneous solutions.
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Example 1
SOLVING A TRIGONOMETRIC
EQUATION BY LINEAR METHODS
Sin θ is negative in quadrants III and IV.
The reference angle is 30° because
Solution set:
{210°,330°}
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Example 2
SOLVING A TRIGONOMETRIC
EQUATION BY FACTORING
Subtract sin θ.
Factor out sin θ.
Zero-factor property
Solution set: {0°, 45°, 180°, 225°}
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Caution
There are four solutions in Example 2.
Trying to solve the equation by dividing
each side by sin θ would lead to just
tan θ = 1, which would give θ = 45° or
θ = 225°. The other two solutions would
not appear. The missing solutions are
the ones that make the divisor, sin θ,
equal 0.
For this reason, avoid dividing by a
variable expression.
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Example 3
SOLVING A TRIGONOMETRIC
EQUATION BY FACTORING
Factor.
Zero-factor property
The solutions for tan x = 1 over the interval
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are
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Example 3
SOLVING A TRIGONOMETRIC
EQUATION BY FACTORING (continued)
Based on the range of the
inverse tangent function,
is in quadrant IV.
Since we want solutions over the interval
to –1.1071487, and then add 2
, add
Solution set:
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Example 4
SOLVING A TRIGONOMETRIC EQUATION
USING THE QUADRATIC FORMULA
Find all solutions of cot x(cot x + 3) = 1.
Write the equation in
standard quadratic form.
Use the quadratic formula with a = 1, b = 3, and
c = –1 to solve for cot x.
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Example 4
SOLVING A TRIGONOMETRIC EQUATION
USING THE QUADRATIC FORMULA
(continued)
Use the reciprocal identity
and
to find the values of x.
To find all solutions, add integer multiples of the
period of tangent, .
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Example 5
SOLVING A TRIGONOMETRIC EQUATION
BY SQUARING
Square both sides.
Expand.
Pythagorean identity
Subtract 3 + tan2 x.
Divide by
, then
rationalize the
denominator.
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Example 5
SOLVING A TRIGONOMETRIC EQUATION
BY SQUARING (continued)
Since the solution was found by squaring both sides
of an equation, we must check that each proposed
solution is a solution of the original equation.
is not a solution.
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is a solution.
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Example 6
DESCRIBING A MUSICAL TONE FROM A
GRAPH
A basic component of music is a pure tone. The graph
below models the sinusoidal pressure y = P in pounds
per square foot from a pure tone at time x = t in
seconds.
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DESCRIBING A MUSICAL TONE FROM A
GRAPH (continued)
Example 6
The frequency of a pure tone
is often measured in hertz.
One hertz is equal to one cycle
per second and is abbreviated
Hz. What is the frequency f in
hertz of the pure tone shown in
the graph?
From the graph, we see that there are 6 cycles in .04
seconds.
The frequency f is 150 Hz.
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Example 6
DESCRIBING A MUSICAL TONE FROM A
GRAPH (continued)
The time for the tone to
produce one complete cycle
is called the period.
Approximate the period T in
seconds of the pure tone.
Six periods cover a time of .04 seconds.
One period =
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Example 6
DESCRIBING A MUSICAL TONE FROM A
GRAPH (continued)
An equation for the graph is
Use
a calculator to estimate all solutions to the equation
the make y = .004 over the interval [0, .02].
The first point of intersection is at about x = .0017 sec.
The other points of intersection are at about x = .0083 sec
and x = .015 sec.
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