10.4 Solve Trigonometric Equations. Solving trig equations… What’s the difference?

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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.
ANSWER
π3 + 2n π or + 2n π5π
6
2 sin x + 4 = 5
2 sin x = 1
sin x = 12
sin1 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.
ANSWER , , ,π3
2π3
4π3
5π3
3 csc2 x = 4
csc2 x = 432
√3csc x =
2√3
sin x =
=sin x1 2
√3
2√3
sin1 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
Solve for t.t = 12.4n=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 AssignmentPage 639, 321 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.
ANSWER The correct answer is D.
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.
ANSWER
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
,
ANSWER π4
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
0, π orANSWER π2
On the interval 0 ≤ x < π, tan x = 0 has two solutions x = 0, π
104 Assignment day 2Page 639, 2435 all