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
To solve the equation 8cos^2 3x - 17cos3x + 3 = 0, we can use the quadratic formula. Let's first rewrite the equation in the form of au^2 + bu + c = 0, where u = cos 3x:
8u^2 - 17u + 3 = 0
Comparing this with the general quadratic equation form, we have a = 8, b = -17, and c = 3. Now we can substitute these values into the quadratic formula:
u = (-b ± √(b^2 - 4ac)) / (2a)
Plugging in the values, we get:
u = (-(-17) ± √((-17)^2 - 4(8)(3))) / (2(8))
u = (17 ± √(289 - 96)) / 16
u = (17 ± √193) / 16
So, cos 3x = (17 ± √193) / 16.
However, the cosine of any angle lies between -1 and +1, so we need to find the values of 3x that give us values within this range.
cos 3x = 1.93.. or cos 3x = 0.194222
To find the values of x, we can use inverse cosine (arccos) to get the angles (in degrees) that have the above cosine values:
3x = arccos(1.93..)
3x = arccos(0.194222)
Using a calculator, we find that the solutions are approximately:
3x = 78.8º or 281.2º
Now, we need to find the values of x by dividing these angles by 3:
x = 26.267º or 93.733º
Lastly, since the cosine function has a period of 2π/3 (or 120º), we can add multiples of 120º to our solutions until we reach values greater than or equal to 360º:
For x = 26.267º and x = 93.733º:
x = 26.267º, 93.733º, 146.267º, 213.733º, 266.267º, 333.733º
Therefore, the solutions to the given equation, to the nearest tenth, where 0 degrees ≤ x < 360 degrees, are:
x = 26.3º, 93.7º, 146.3º, 213.7º, 266.3º, and 333.7º.