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# Trig Q watch

1. Can someone tell me how to approach this question (its in AEA 04 )

1) cosx + √ (1-0.5sin2x) = 0

Thanks!
2. what's the domain?
3. (Original post by darkenergy)
Can someone tell me how to approach this question (its in AEA 04 )

1) cosx + √ (1-0.5sin2x) = 0

Thanks!
You could square both sides but crucially you need to remember you lose the signs.
Alternatively you could multiply both sides by to give:

but you need to remember
4. I just plotted the graph, and only get 1 solution...2
How is this possible, if there are 3 terms in the eq?
5. (Original post by Goldenratio)
I just plotted the graph, and only get 1 solution...2
How is this possible, if there are 3 terms in the eq?
It's defined over (0,360). It's not a polynomial so I don't think there are any explicit rules about how many roots it might have that we should know of.
I get 2 roots. Try solve it algebraically.
6. what if the equation becomes

cosx + √(1-0.5sin2x) + 2 = 0

is this solvable?
7. (Original post by darkenergy)
what if the equation becomes

cosx + √(1-0.5sin2x) + 2 = 0

is this solvable?
I don't think that has solutions.
You're saying: cosx+√(1-0.5sin2x)=-2
but √(1-0.5sin2x)>=0 so you need cosx<=-2.
But |cosx|<=1 so no solutions exist.
8. what if it is:

cosx+√(1-0.5sin2x)= 2?

I am just wondering how to solve such equations when there is a number term so you can just square both sides
9. I started with your lead but i can't get rid of the -1.
First so i did:
cos^2x-(1-1/2sin2x)=0
=> cos^2x-1+sinxcosx=0
therefore
cosx(cosx+sinx)=1
cosx+sinx=secx
square both sides so: cosx^2+2sinxcosx+sinx^2=sec^2x
1+sin2x=sec^2x
2sinxcosx=tan^2x
sin2x=tan^2x but then realised it just complicated it...
10. (Original post by darkenergy)
what if it is:

cosx+√(1-0.5sin2x)= 2?

I am just wondering how to solve such equations when there is a number term so you can just square both sides
Let's try:
√(1-0.5sin2x)=(2-cosx).
Evidently both sides are greater than 0 for all x so you can square without adding spurious solutions.
1-0.5sin2x=4-4cosx+(cosx)^2
(cosx)^2-4cosx+3+0.5sin2x=0
(cosx)^2-4cosx+3+sinxcosx=0
c^2-4c+3+c√(1-c^2)=0
It clearly has solutions (eg x=0) but it's not easy to solve exactly.
11. I'd do it in the following way:
12. Ah! yep nice one!
13. k thanks people

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