1. Aim: How do we solve first and second degree trig equation?
Do Now:
Solve for x: 6x + 7 = 10
2. Given the equation 6sin x + 7 = 10
find:
a) sin x b) x in the domain
HW: p.524 # 6,8,12,14, p # 8,12, p.534 # 6,8

2. 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.

3. Make cos θ and number on each side of equation
Given
find all values of
in the interval of
Make cos θ and number on each side of equation
Solve for cos θ
or 240
Solve for θ

4. Solve for 2. Solve for 3. Solve for in the interval : in the interval

5. Solve for x over
Treat sin2 x like x2 , and factor the trinomial into two binomials
(2sin x + 1)(sin x – 3) = 0
Set each binomial = 0,and solve for θ
2sin x + 1 = 0, sin x – 3 = 0
x = 210 or 330
reject

6. Find all the values of θ in the interval
Factor the trinomial is always the first choice, If the equation is not factorable, then we need to use quadratic formula

7. 1.707
0.293
Done

8. reject
The value of cos θ can not be > 1
θ is in quadratic I or IV since cos θ is positive
In first quadrant
In fourth quadrant

9. 1. Solve the equation: 2. Solve the equation: 3. Solve the equation:
for all values of in the interval
(60, 90, 270, 300)
2. Solve the equation:
for all values of in the interval
3. Solve the equation: (
for all values of in the interval
4. Solve the equation:
for all values of in the interval
(205.44, )