Aim: How do we solve first and second degree trig equation? Do Now: 1. Solve for x: 6x + 7 = 10 2....

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Aim: How do we solve first and second degree trig equation? Do Now: 1. Solve for x: 6x + 7 = 10 . 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

Transcript of Aim: How do we solve first and second degree trig equation? Do Now: 1. Solve for x: 6x + 7 = 10 2....

Page 1: 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.

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

Page 2: 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.

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.

Page 3: 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.

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

Page 4: 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.

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 :

Page 5: 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.

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

Page 6: 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.

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

Page 7: 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.

)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

Page 8: 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.

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

Page 9: 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.

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)