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trig

posted by .

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.

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