Trig Equations (1 Viewer)

[tWiStEdD]

i've tried my hardest, and cannot for the life of me find the answers to the following questions:

5 sin x = 2 sec x

and

cos 2x * cos (pi/6) - sin 2x * sin (pi/6) = 1/2

hrm... what methods should i use?

 

[Xayma]

The first one it is probably easiest to use ratios in terms of tan theta/2.

 

[tWiStEdD]

i tried that, but came out with:

2t-2(t^3)
--------------- = 2/5
(1-(t^2))^2
what've i done wrong???

 

[KeypadSDM]

5Sin[x]=2Sec[x]
2.5Sin[x]Cos[x]=1
1.25Sin[2x]=1
Sin[2x] = 4/5

And so on.

cos 2x * cos (pi/6) - sin 2x * sin (pi/6) = 1/2
Cos[2x + pi/6] = 1/2

And so on.

 

[tWiStEdD]

w00t maths god!

 

[KeypadSDM]

Originally posted by tWiStEdD
i tried that, but came out with:

2t-2(t^3)
--------------- = 2/5
(1-(t^2))^2
what've i done wrong???

You haven't done anything wrong. But you haven't factorized completely.

Continuing

1/5 = t(1 - t^2)/(1 - t^2)^2
1/5 = t/(1 - t^2)
0 = t^2 + 5t - 1

Etc.

(I don't know if the first line is right tho, couldn't be bothered checking )