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Quick Question:
Is 2cos^2 x+2sin^2x=2?
Is 2cos^2 x+2sin^2x=2?
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Yes
2(cos^2 + sin^2)
2(1)
2
Thankyouu soo much i got it nowfrom solving first part you are able to get to sinx=0.5 . now if you compare both equations the only diff is n so we can say that
30/10=n hence n= 3 now for greatest solution find all solutions up to 3 x 360. then divide the highest with 3 you will get the answer
sin 3x + 2cos 3x = 0http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s03_qp_1.pdf
Please help me with question 2, I don't know what to do.
Thanks
you can substitute 3x to something else like y (anything) so 3x = ysin 3x + 2cos 3x = 0
divid the equation by cos x and u will get tan 3x +2 = 0
tan 3x = -2
3x = -64.3 and 116.5
x= 158.9 and 38.9 but i have no idea how the mark scheme got 98.9 as one of x values sorry :/
Silent Hunterok here goes to find the limits of this curve y= sin cube 2x cos cube 2x we will put y=0 as on the x axis y=o
0= cos cube2x sin cube 2x o=sin cube 2x
sin cube 2x = 0 0 = cos cube 2x
x=0 2x=pi/2
x= pi/4
so the limits be pi/4 and 0
by substituting u= sin2x
du/dx = 2cos2x
du= 2cos2x dx
now substituting
(cos cube 2x)( u^3) (1/ 2cos2x) <----------- where this is du
one cos would be cut by the cos in the denominator
u^3 (cos square 2x)/2
remember that cos square x = 1-sin square x
so cos square x = 1-sin square x
so
cos square 2x = 1-sin square 2x
and sin 2x =u
so
cos square 2x= 1- u^2
then the total thing wud be
1/2 integral of u^3(1-u^2)
1/2 integral of u^3 - u^5
1/2 (( u^4)/4 - (u^6)/6)
now open u as sin2x
1/2 (( sin 2x^4)/4 - (sin 2x^6)/6)
1/2 (1/4- 1/6)
1/2 (1/12)
1/24
No need to thank me
because i didn't solve it
Rutzaba solved it for me
oh that shows how to get it, thank youyou can substitute 3x to something else like y (anything) so 3x = y
so you have tan y =-2, you have to change the range 0<x<180 to 0<y(which is 3x)<540
then you have y= -63.4 + 180 (keep adding 180)
y2= 116.6 and you put it back into y =3x which gives you 38.9
y3= 296.6 => 98.9
y4=476.6 => 158.9
find f inverse(x) and put x^2-2 in place of x@@@@@@
If f(x) > 2x+1
and g(x) > x^2-2
find f^-1 g(x) ??? ::::::
got ithttp://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w09_qp_31.pdf
N0 10i Dug PhyZac or any1 having p3 please help
da/dt = KVhttp://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w09_qp_31.pdf
N0 10i Dug PhyZac or any1 having p3 please help
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