Trigonometry/ProOneSoftware-Trigonometry-Win31.iso/trig/chapter4.7r

Text File | 1995-04-09 | 5KB | 162 lines

 160
à 4.7ïTrigonometric Equations

äïPlease find the primary solutions of the following
êêequations.
âêFind the primary solutions of 2∙sin x - 1ï=ï0.

êêêë 2∙sin x - 1ï=ï0
êêêê sin xï=ï1/2
#êêêêxï=ïSinúî(1/2)
êêêêxï=ïπ/6, 5π/6
éSïIn the previous sections of this chapter, we have been working
with identities.ïThese, as you know, are equations that are true for
all angles or all values of x when substituted into the equation.ïIn
this section, we would like to look at solving equations that are true
only for certain values of x.ïAn equation that is true only for cer-
tain values of x is called a conditional equation.
è In the example, you are asked to solve the conditional equation,
#2∙sin x - 1ï=ï0.ïThe first step is to solve for x.ïxï=ïSinúî(1/2).
#Next, you should use your calculator to find Sinúî(1/2).ïYou will get
a reference angle of 30° or π/6.ïSince the sin x is positive in the
first and second quadrants, the solutions to the equation are π/6 and
5π/6.ïSince the sin x is periodic, there are many other solutions to
this equation.ïAny multiple of 2π added to either of ç solutions is
also a solution.ïThus, π/6 + n∙2π and 5π/6 + n∙2π are all of the solu-
tions of the equation.ïUsually, you will need only those solutions
that fall between 0° and 360°.ïThese are the primary solutions of the
equation.ïIn the example, the primary solutions of 2∙sin x - 1ï=ï0
are π/6 and 5π/6.
è Another way to approach this problem is to return to the Key Fea-
ture in this program.ïThere you will seek those vallues of x for which
the y coordinate of P(x) is 1/2.ïThere are only two points on the unit
circle with y coordinate 1/2.ïThey are 30° and 150° or π/6 and 5π/6.
Thus, once again the solutions to the equation, 2∙sin x - 1ï=ï0, are
π/6 and 5π/6.ïYou can also draw a graph of y = sin x on your Function
Plotter and see that when y is 1/2, x is either 30° or 150° in the inter-
val from 0° to 360°.
è There are several techniques used at different times to solve trigo-
nometric equations.ïThese techniques include factoring, multiplying by
the LCD, using identities to simplify, using the quadratic formula, and
squaring both sides of the equation.ïIt is necessary to check for ex-
traneous solutions, however, when you square both sides.
 1êëFind the primary solutions of
êêêê2∙cos xï=ï√3

ê A)ïπ/6, 5π/6êêêèB)ïπ/6, 11π/6

ê C)ïπ/6, 7π/6êêêèD)ïå of ç
ü
êêêê2∙cos xï=ï√3

#êêêêxï=ïCosúî(√3/2)
êêëThe reference angle is x = π/6 or 30°.
Since cos x is positive in quadrants I and IV, the primary solutions are
π/6 and 11π/6.
Ç B
 2êëFind the primary solutions of
êêêêtan xï=ï- √3

ê A)ï2π/3, 5π/3êêêïB)ïπ/3, 5π/3

ê C)ïπ/3, 4π/3êêêèD)ïå of ç
ü
êêêêtan xï=ï- √3

#êêêêxï=ïTanúî(-√3)
êêëThe reference angle is x = -π/3 or -60°.
Since tan x is negative in quadrants II and IV, the primary solutions
are 2π/3 and 5π/3.
Ç A
 3êëFind the primary solutions of
êêêêcot xï=ï- 1

ê A)ïπ/4, 5π/4êêêèB)ï3π/4, 5π/4

ê C)ï3π/4, 7π/4êêêïD)ïå of ç
ü
êêêêcot xï=ï- 1

#êêêêxï=ïCotúî(-1)
êêëThe reference angle is x = -π/4 or -45°.
Since cot x is negative in quadrants II and IV, the primary solutions
are 3π/4 and 7π/4.
Ç C
 4êëFind the primary solutions of
êêêè sin 2x + cos xï=ï0

ê A)ïπ/2, π/6êêêè B)ïπ/4, π/6

ê C)ïπ/2, 3π/2, 7π/6, 11π/6êè D)ïå of ç
ü
êêêësin 2x + cos xï=ï0

êêêï2∙sin x ∙ cos x + cos xï=ï0

êêêè cos x (2∙sin x + 1)ï=ï0

#êêêïcos xï=ï0 │ 2∙sin x + 1ï=ï0
#êêëxï=ïCosúî(0)ï│ xï=ïSinúî(-1/2)
#Reference angles,ïx = π/2 or 3π/2 │ x = -π/6 or -30°
Since sin x is negative in quadrants III and IV, the primary solutions
are π/2, 3π/2, 7π/6, and 11π/6.
Ç C
 5êëFind the primary solutions of
#êêêï2∙sinìx + sin x - 1ï=ï0

ê A)ïπ/6, 5π/6, 3π/2êêè B)ï5π/6, π/2

ê C)ï7π/6, 11π/6, π/2êêèD)ïå of ç
ü
#êêêè2∙sinìx + sin x - 1ï=ï0

êêê (2∙sin x - 1)(sin x + 1)ï=ï0
#êêêêê │
#êêè 2∙sin x - 1 = 0è│ïsin x + 1 = 0
#êêêêê │
#êêëx = Sinúî(1/2)è│ïx = Sinúî(-1)
#Reference angles,ïx = π/6êè│ïx = 3π/2
#Since sin x is positive in quadrants│ Since sin x is negative one only
#I and II, the primary solutions are │ at 3π/2 in the interval 0° to
#π/6, 5π/6.êêêë│ 360°, it is the primary solution.
Ç A
 6êëFind the primary solutions of
#êêê 2∙sinìx - 2∙sin x - 1ï=ï0

ê A)ï2.8759, 1.7326êêëB)ï3.5163, 5.9087

ê C)ï2.4712, 4.7492êêëD)ïå of ç
üêêèUse the quadratic formula.
#êêêï2∙sinìx - 2∙sin x - 1ï=ï0
#êêêêè ┌───────────
#êêêê2 ± á4 - 4∙2(-1)ë1 ± √3
#êêè sin xï=ï────────────────ï=ï──────
êêêêê 2∙2êë2
êêêïsin xï=ï-.3660 or 1.366
1.366 is rejected, because it is outside of the range of sin x.
#êêê Thus, xï=ïSinúî(-.3660).
Your calculator can be used to find a reference angle of -.3747.ïSince
sin x is negative in quadrants III and IV, the primary solutions are
3.5163 and 5.9087.
Ç B
 7êëFind the primary solutions of
êêêè2∙cos x - cot xï=ï0

ê A)ïπ/2, π/6, π/4êêë B)ïπ/4, 5π/4

ê C)ïπ/2, 3π/2, π/6, 5π/6êë D)ïå of ç
ü
êêêè2∙cos x - cot xï=ï0

êêêï2∙cos x - cos x/sin xï=ï0

êêê 2∙sin x ∙cos x - cos xï= 0

êêêè cos x (2∙sin x - 1) = 0
#êêê cos x = 0ë│ï2∙sin x - 1 = 0
#êêêêê │ïx = Sinúî(1/2)
#êêêêê │ïReference angle is π/6.
#êêë x = π/2 or 3π/2 │ïPrimary solutions are π/6, 5π/6.
Ç C