April 1, 2015

Homework Help: precal

Posted by justy on Monday, May 14, 2007 at 9:53pm.

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 = 78.8 or 3x = 281.2

therefore x = 26.267 or x = 93.733

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

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

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

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