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.

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Aim: How do we solve first and second degree trig equation? Do Now: 1. Solve for x: 6x + 7 = Given the equation 6sin x + 7 = 10 find : a) sin x b) x in the domain 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 find all values of in the interval of Make cos θ and number on each side of equation Solve for cos θ Solve for θ or 240 

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

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

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

 Done  0.293

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

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