Question

how to find the solution in the interval of 0 to 360 deg for 2cos u+cos 2u=0

Answer 1

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Answer 2

I think if you make a substitution using the identity:

\(\Large \cos2u=2\cos^2u-1\)

Then you will have an equation of quadratic type and can solve with quadratic equation methods.

Answer 3

i have done that
but it is not real number

Answer 4

You should get 2 solutions for cos(u).... then use the inverse function to find u.

Can you show your work to solve the quadratic? You should get 2 irrational solutions.

Answer 5

yep i get two irrational solution
from that how do i do next

Answer 6

\[\Large 2\cos u+\cos 2u=0\]\[\Large 2\cos u+2\cos^2u-1=0\]
disriminant =\(\large b^2-4ac=4-4(2)(-1)=4+8=12\) so should have 2 irrational solutions.

Answer 7

OK, remember that those are solutions for cosu.

So you have cos(u)=a

You want u. and you know the cosine of u. What "tool" (think of a function that you can use) takes the COSINE VALUE of an angle as the input, and gives you the ANGLE MEASURE as the output?

Answer 8

These are your solutions for the quadratic?

\[\Large \cos u=\frac{ -1\pm \sqrt{3} }{ 2}\]

Answer 9

oh ok
so the two solution is the exact value of cos u
ok ok i got it
thank u

Answer 10

you're welcome. :)

Answer 11

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