trigonometric equations 5.5. to solve an equation containing a single trigonometric function:...
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Trigonometric Equations
5.5
To solve an equation containing a single trigonometric function:
• Isolate the function on one side of the equation.
• Solve for the variable.
Equations Involving a Single Trigonometric Function
y = cos x
x
y
1
–1
y = 0.5
–4 2–2 4
cos x = 0.5 has infinitely many solutions for – < x <
y = cos x
x
y
1
–1
0.5
2
cos x = 0.5 has two solutions for 0 < x < 2
Trigonometric Equations
Solve the equation: 3 sin x 2 5 sin x 1.
Solution The equation contains a single trigonometric function, sin x.
Step 1 Isolate the function on one side of the equation. We can solve for sin x by collecting all terms with sin x on the left side, and all the constant terms on the right side.
3 sin x 2 5 sin x 1 This is the given equation.
3 sin x 5 sin x 2 5 sin x 5 sin x – 1 Subtract 5 sin x from both sides.
sin x -1/2
Divide both sides by 2 and solve for sin x.
2 sin x 1 Add 2 to both sides.
2 sin x 2 1 Simplify.
Text Example
Solve the equation: 2 cos2 x cos x 1 0, 0 x 2.
The solutions in the interval [0, 2) are /3, , and 5/3.
Solution The given equation is in quadratic form 2t2 t 1 0 with t cos x. Let us attempt to solve the equation using factoring.
2 cos2 x cos x 1 0 This is the given equation.
(2 cos x 1)(cos x 1) 0 Factor. Notice that 2t2 + t – 1 factors as (2t – 1)(2t + 1).
cos x 1/2
2 cos x 1 cos x 1 Solve for cos x.
2 cos x 1 0 or cos x 1 0
Set each factor equal to 0.
Text Example
x x 2 x
Example
cos29cos7 • Solve the following equation:
Solution:
n2
5,3,
1cos
9cos9
cos29cos7
Example• Solve the equation on the interval [0,2)
Solution:3
3
2tan
3
7
3
6
7
62
3
3
2tan
and
and
8 sin = 3(1 sin2 ) Use the Pythagorean Identity.
Rewrite the equation in terms of only one trigonometric function.
Example: Solve 8 sin = 3 cos2 with in the
interval [0, 2π].
3 sin2 + 8 sin 3 = 0. A “quadratic” equation with sin x as the variable
Therefore, 3 sin 1 = 0 or sin + 3 = 0
(3 sin 1)(sin + 3) = 0 Factor.
Solutions: sin = or sin = -31 3
= sin1( ) = 0.3398 and = π sin1( ) = 2.8107.1 3
1 3
Example• Solve the equation on the interval [0,2)
Solution:03cos2cos2 xx
0
0
1cos3cos
01cos03cos
0)1)(cos3(cos
03cos2cos2
x
xsolutionno
xx
xx
xx
xx
Example• Solve the equation on the interval [0,2)
Solution:
3
5,
3
2
1cos
1cos2
sincossin2
sin2sin
x
x
x
xxx
xx
xx sin2sin
NO!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
!!!!!!!!!!!!!!!!!