3 cos x = 5-4 sin x
9 cos^2 x = 25 - 40 sin x + 16 sin^2 x
9 (1 -sin^2x) = 16 sin^2 x -40 sin x + 25
9 - 9 sin^2 x = 16 sin^2 x -40 sin x + 25
0 = 25 sin^2 x - 40 sin x + 16
let z = sin x
25 z^2 - 40 z + 16 = 0
(5 z - 4)(5 z - 4) = 0 wow, that was lucky
sin x = 4/5 :)
3cosx+4sinx=5
what is x is too difficulat form me to solve?
Shw working please
6 answers
Im not in Trig yet but i did look up online trigonometry calculators... see if that help?
PS, got this from 1 of those sites:
3cos(x) + 4sin(x) = 5 : x = 2πn + arcsin(4/5)
Hope this helps.
PS, got this from 1 of those sites:
3cos(x) + 4sin(x) = 5 : x = 2πn + arcsin(4/5)
Hope this helps.
let's change 3cosx+4sinx into a single sine curve of the form y = a sin (x+k)
we know asin(x+k)
= asinxcosk + acosxsink
comparing our terms
asinxcosk = 4sinx
acosk = 4
a = 4/cosk
acosxsink = 3cosx
a sink = 3
a = 3/sink
then 3/sink = 4/cosk
4sink = 3cosk
sink/cosk = 3/4
tank = 3/4
k = 36.87°
also a^2 = 3^2 + 4^2
a = 5
so 3cosx+4sinx=5 becomes
5sin(x+36.87°) = 5
sin(x+36.87°) = 1
x+36.87° = 90°
x = 53.13°
check:
LS = 3cos53.13 + 4sin53.12
= 4.99958
not bad, we could carry more decimals for more accuracy
of course we could get more solutions by simply adding multiples of 360° to 53.12°
e.g x = 413.13 or x = -306.87° etc
we know asin(x+k)
= asinxcosk + acosxsink
comparing our terms
asinxcosk = 4sinx
acosk = 4
a = 4/cosk
acosxsink = 3cosx
a sink = 3
a = 3/sink
then 3/sink = 4/cosk
4sink = 3cosk
sink/cosk = 3/4
tank = 3/4
k = 36.87°
also a^2 = 3^2 + 4^2
a = 5
so 3cosx+4sinx=5 becomes
5sin(x+36.87°) = 5
sin(x+36.87°) = 1
x+36.87° = 90°
x = 53.13°
check:
LS = 3cos53.13 + 4sin53.12
= 4.99958
not bad, we could carry more decimals for more accuracy
of course we could get more solutions by simply adding multiples of 360° to 53.12°
e.g x = 413.13 or x = -306.87° etc
Go with Damon's , he used an easier method
(but I thought mine was quite elegant)
(but I thought mine was quite elegant)
well I agree with anonymous :)
thnx Damon. im not in trig so i just looked it up :)