# precal

posted by
**justy**
.

when i do the quadratic equation i get -17 plus or minus the square root of 193 all over 2. where do i get the two "cos 3x" roots??

solve the equation to the nearest tenth, where 0 degrees < or equal to x which is < 360 degrees.

8cos^2 3x-17cos3x+3=0

this doesnt factor, when i used the quadratic formula i got -17 plus or minus the square root of 193 over 2?? is that right?

Let the variable be u = cos 3x

8 u^2 -17 u +3 = 0

Use the quadratic equation to get the two "cos 3x" roots. Then use trig to get 3x

not quite

your solution would be

cos 3x = (17 ± sqrt(193))/16

cos 3x = 1.93.. or cos 3x = .194222

but the cosine of any angle lies between -1 and +1 so

cos 3x = .194222

then 3x = 78.8º or 3x = 281.2º

therefore x = 26.267º or x = 93.733º

Now lastly:

The period of cos 3x, which was the original trig function we were dealing with, is 360/3º or 120º

so other answers are obtained by adding 120 to each of my previous answers until we run over 360º

so x = 26.267, 93.733, 146.267,213.733, 266.267, 333.733