Hey all,
I've been trying very hard to figure out this particular question for quite some time now already. Its gotta do with the R-formulae in further trigo.
Hope anyone here can advice me on how to solve it, cause even my math teacher cant
So here it is:
Solve the following equations for 0< x<360
2 sin x (2 sin x - cos x) = 1
Thanks =)
2 sin x (2 sinx - cos x) = 1
4 sin^2 x - 2sin x cos x = 1
4[1/2-1/2 cos (2x)] - sin (2x) =1
2- 2cos (2x) - sin (2x) =1
2 cos (2x) + sin (2x) = 1
Let 2 cos (2x) + sin (2x) = R cos (2x-P)
R = sqrt 5 and tan P = 1/2---> P = 26.56 degree
Therefore sqrt (5) cos (2x - 26.56) =1
cos ( 2x - 26.56) = 1/(sqrt 5)
Edited due to carelessness.
I believe u can continue from here.
Feel free to ask all your A Maths questions here. The tutors here enjoy helping people. Personally, for myself, it is because I wish my students will ask me questions... but they seldom do. :(
Hey thanks a lot!
Didn't expect a respond that fast, it really helped very much! =)
Really cool place to clear my doubts.
However, I don't understand from step 1 to step 2.
"2 sin x (2 sinx - cos x) = 1
4 sin^2 x + 2sin x cos x = 1"
Shouldn't it be 4sin^2x - 2sin xcos x = 1 ?
And also from step 2 to 3.
And from
2- 2cos (2x) + sin (2x) =1
to
2 cos (2x) - sin (2x) = 1 , also don't make sense to me..
Thanks
step 1-2 is just multiplying 2sinx to the terms in the bracket...
From 4-5 intermediate steps are
2-2cos 2x + sin 2x =1
-2cos 2x + sin 2x = -1 (-2 on both sides)
2cos 2x - sin 2x = 1 (multiply by -1 on both sides)
I'm not a tutor by the way =)
For Step 2-3, it's all double angle formula
4 sin^2 x + 2sin x cos x = 1
separating the 2 terms,
cos 2x = 1 - 2sin^2 x 2sinx cosx = sin 2x
2sin^2 x = 1 - cos 2x
sin^2 x = 1/2 - 1/2cos 2x
So...
4[1/2-1/2 cos (2x)] + sin (2x) =1
Alright thanks lots!
Managed to get the answers right already xD
I see the sign wrongly mah. :P
coz I looking on my smaller monitor as I was installing Windows 7 on my main rig. :P
Will edit my post.