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 3x17cos3x+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
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