it says solve for all real solutions, compute inverse functions to 4 sigfigs.
cos^2x=3-5cosx
Devon, what grade level are these questions from?
I am having the same problems as with your 9:14 post.
Please see my partial solution in the post to bobpursely at 10:54
Devon, I have been misreading your questions, yes, they are readily solvable.
cos^2(x) + cosx -3 = 0, treat it as a quadratic with cosx the variable
I got cosx = .54138 or cosx = -5.5...
The last one is not possible so we have to use
cosx = .54138
using arccos gives us 57.2º
Now recall that the cosine in positive in the first and fourth quadrants, so
x=57.2º or 360º-57.2º = 302.8º
these answers are in the domain 0<x<360, if you want a general answer it would be
x = 57.2 + 360n or 302.8 +360n where n is an integer.
For radians, set your calculator to RAD and follow the above method
For general solutions add 2(pi)n to each answer you found
thank you so much, i totally understand it now!
to your grade level question, i'm in 9th grade, but i'm pretty sure it's 11th grade material.
I'm glad to hear that you understand the solution now! Solving the equation cos^2x = 3 - 5cosx involves finding the values of x in the given domain for which the equation is true.
To solve for all real solutions, you correctly converted the equation to a quadratic equation in terms of cosx: cos^2x + cosx - 3 = 0. Then, you factored it as (cosx - 1)(cosx + 3) = 0.
Since the product of two factors is zero, either cosx - 1 = 0 or cosx + 3 = 0. Solving each equation separately gives cosx = 1 or cosx = -3.
Considering the range of the cosine function (-1 ≤ cosx ≤ 1), the solution cosx = -3 is not possible. Therefore, you only use the solution cosx = 1.
Using the inverse cosine function (arccos), you found cos^-1(1) = 0°. This gives you one solution: x = 0°.
Next, you considered the periodic nature of the cosine function. Since cosx repeats every 360° in the domain 0° < x < 360°, you added multiples of 360° to your initial solution.
So, the general solution is x = 0° + 360°n or x = 360° - 0° + 360°n, where n is an integer. This accounts for all possible solutions in the given domain.
If you want the solution in radians, you can set your calculator to RAD and follow the same method. The general solution in radians would be x = 0 + 2πn or x = 360° - 0 + 2πn, where n is an integer.
Remember to compute inverse functions with 4 significant figures as mentioned in your question.
Regarding the grade level, it seems that the material you are studying is more advanced and typically covered in 11th grade or higher. It's great that you are tackling challenging topics and seeking a deeper understanding!