Post on 18-Jan-2016
Aim: How do we solve first and second degree trig equation?
Do Now:
1. Solve for x: 6x + 7 = 10
2. Given the equation 6sin x + 7 = 10 find:
a) sin x b) x in the domain 0 360 x
HW: p.524 # 6,8,12,14,16 p.530 # 8,12,14 p.534 # 6,8
6 sin x = 3, sin x = 1/2,
since sin x is positive, we know we’re in quadrant I and II so x = 30 and also150
To solve for the tirg equation of x is actually no different from solving for x in a regular equation.
solve for x itself is simply a matter of 1) determining what quadrant you are working in and from that 2) what angles x equals by using the inverse function and consider also the given interval.
Given cos cos 3 1
0 2 find all values of in the interval of
Make cos θ and number on each side of equation
1cos2
Solve for cos θ2
1cos
Solve for θ
1202
1cos 1 or 240
0 360 2 3 0sin
0 2 3 3cos cos
0 2 2 1 3(sin ) sin
1.Solve for in the interval
:
2. Solve for in the interval
:
3. Solve for in the interval :
0 360 x :2 5 3 02sin sinx x
Solve for x over
(2sin x + 1)(sin x – 3) = 0
Treat sin2 x like x2 , and factor the trinomial into two binomials
Set each binomial = 0,and solve for θ
2sin x + 1 = 0, sin x – 3 = 0
2
1sin
x
x = 210 or 330
3sin x
reject
01cos4cos2 2 Find all the values of θ in the interval 3600
Factor the trinomial is always the first choice, If the equation is not factorable, then we need to use quadratic formula
)2(2
)1)(2(4)4()4( 2 x
4
8164
4
84
2
22
)2(2
)1)(2(4)4()4(cos
2
4
8164
4
84
2
22
1.207
Done
2
22
2
22 0.293
207.1cos reject
The value of cos θ can not be > 1
293.0cos θ is in quadratic I or IV since cos θ is positive
73
28773360
In first quadrant
In fourth quadrant
2 2cos cos 0 360
1. Solve the equation: for all values of in the interval
xx cos2cos3 0 2
2. Solve the equation:
for all values of in the interval
sin sin2 0 0 360
3. Solve the equation:
for all values of in the interval
10 12 7 02sin sinx x 0 360
4. Solve the equation:
for all values of in the interval
(60, 90, 270, 300)
(
/ , , / )3 5 3
( , / , )0 2
(205.44, 334.56)