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10.4 Solve Trigonometric Equations

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### Transcript of 10.4 Solve Trigonometric Equations. Solving trig equations… What’s the difference? 10.4 Solve Trigonometric Equations Solving trig equations…What’s the difference? Solve 2 sin x – 3 = 0.

SOLUTION

First isolate sin x on one side of the equation.

Write original equation. 2 sin x – 3 0=

Add 3 to each side.2 sin x 3=

Divide each side by 2.sin x 32

=

One solution of sin x = 32 in the interval 0 ≤ x < 2 π is

x 32= sin–1 π

3 .=

The other solution in the interval is x π3–= π 2π

3 .=

Solve a trig equation  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:

x + 2nπ π3= or x + 2nπ 2π

3= (where n is any integer)

CHECK Solve a trigonometric equation in an intervalSolve 9 tan2 x + 2 = 3 in the interval 0 ≤ x <2π.

Write original equation.9 tan2 x + 2 3=

Subtract 2 from each side.9 tan2 x 1=

Divide each side by 9.tan2 x19=

Take square roots of each side.tan x 13

+–=Using a calculator, you find that

tan –1 13

0.322 and 13

tan –1 (– ) – 0.322.

Therefore, the general solution of the equation is:

x or0.322 + nπ 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 2.820x 0.322 + π 3.464 x – 0.322 + 2π 5.961 1. Find the general solution of the equation 2 sin x + 4 = 5.

π3 + 2n π or + 2n π5π

6

2 sin x + 4 = 5

2 sin x = 1

sin x = 12

sin-1 x = π3

5π6

,

Write original equation.

Subtract 4 from each side.

Divide both sides by 2.

Use a calculator to find both solutions within 0 < x < 2π 2. Solve the equation 3 csc2 x = 4 in the interval 0 ≤ x <2π.

2π3

4π3

5π3

3 csc2 x = 4

csc2 x = 432

√3csc x =

2√3

sin x =

=sin x1 2

√3

2√3

sin-1 x =

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 Oceanography

The water depth d for the Bay of Fundy can be modeled by

d π6.2

t= 35 – 28 cos

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.

Substitute 7 for d. 35 – 28 cos π6.2

t 7=

Subtract 35 from each side.– 28 cos π6.2

t –28=

Divide each side by –28.π

6.2t cos 1=

cos q = 1 when q = 2nπ.π6.2

t = 2nπ

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 AssignmentPage 639, 3-21 all 10.4 Solve Trigonometric Equations SOLUTION Write original equation.sin3 x – 4 sin x 0=

Factor out sin x.sin x (sin2 x – 4) 0=

Factor difference of squares.sin x (sin x + 2)(sin x – 2) 0=

Set each factor equal to 0 and solve for x, if possible.

sin x = 0x = 0 or x = π

sin x + 2 = 0sin x = –2

sin x – 2 = 0sin 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. Eliminate solutionsBecause 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 formulaSolve cos2 x – 5 cos x + 2 = 0 in the interval 0 ≤ x <π.

SOLUTION Because the equation is in the form au2 + bu + c = 0 where u = cos x, you can use the quadratic formula to solve for cos x.

Write original equation.cos2 x – 5 cos x + 2 = 0

Quadratic formulacos x =– (–5) + (–5)2 – 4(1)(2)

2(1)–

Simplify.5 + 17

2–=

Use a calculator. 4.56 or 0.44

Use inverse cosine.x = cos –1 0.44

Use a calculator, if possibleNo solution 1.12

ANSWER In the interval 0 ≤ x ≤ π, the only solution is x 1.12. Solve an equation with an extraneous solutionSolve 1 + cos x = sin x in the interval 0 ≤ x < 2π.

Write original equation.1 + cos x sin x=

Square both sides.(1 + cos x)2 (sin x)2=

Multiply.1 + 2 cos x + cos2 x sin2 x=

Pythagorean identity1 + 2 cos x + cos2 x 1– cos2 x=

Quadratic form2 cos2 x + 2 cos x 0=

Factor out 2 cos x.2 cos x (cos x + 1) = 0

Zero product property2 cos x = 0 or cos x + 1 = 0 cos x = 0 or cos x = –1 Solve for cos x.On the interval 0 ≤ x <2π, cos x = 0 has two solutions:

x π2= orx 3π

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:

x π2= , π, and

3π2

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

1 + cos π2 = sinπ

2? 1 + cos π ?

= sin π 3π21 + cos 3π

2?= sin

1 + 0 ?= 1 1 + (–1)?= 0 1 + 0

?= –1

1 = 1 0 = 0 1 = –1ANSWER Graphs of each side of the original equation confirm the solutions. Find the general solution of the equation.

4. sin3 x – sin x = 0

sin x(sin2 x – 1) = 0

sin x(– cos2 x ) = 0

sin x = 0 OR (– cos2 x ) = 0

Write original equation. sin3 x – sin x = 0

Factor

Pythagorean Identity

Zero product property

Solvesin x = 0 OR cos x = 0

0 + n π or + n πANSWER π2

On the interval 0 ≤ x <2π, cos x = 0 has two solutions:

x = ,π2

3π2

On the interval 0 ≤ x <2π, sin x = 0 has two solutions:x = 0, π. Find the general solution of the equation.

5. 1 – cos x = sin x31 – cos x = sin x31 – 2 cos x + cos2 x = 3 sin2 x1 – 2 cos x + cos2 x = 3 (1 – cos2 x)

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

cos x = OR cos x = 112

Write original equation.Square both sides.

Pythagorean Identity

Distributive Property

Group like termsFactor

Factor

Zero product property

Solve for cos x 2π3

0 + 2n π or + 2n πANSWER

On the interval 0 ≤ x < 2π, cos x = 1 has one solution:

x = 0.

122π

3

On the interval 0 ≤ x < 2π, cos x = – has two solutionsx = 4π

3

But sine is negative in Quadrant III, so is not a solution4π3 Solve the equation in the interval 0 ≤ x < π.6. 2 sin x = csc x

2 sin x = csc x

2 sin x = sin x1

2 sin2 x = 1

sin2 x = 21

sin x = 2√2

Write original equation.

Reciprocal identity

Multiply both sides by sin x

Divide both sides by 2

Take the square root of both sides

π4

3π4

,

3π4

, 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  