tan 225 = tan 45 = 1; therefore
cos2x = 1/2
2x = 60 or 300 degreees
x = 30 or 150 degrees
I can't do the other one
solve for
2cos2x=tan225
and,
solve using transformation method
cosx+ root2sinx= root3
THANKS!
9 answers
for the first one, since we know tan 225 = 1
your equation is
2cos2x = 1
cos2x = 1/2
2x = 30 degrees or 2x = 150 degrees
x = 15 or x = 75 degrees
you state no domain, but since the period of cos2x is 180 degrees, more answers can be obtained by simply adding/subtracting multiple of 180 to the above answers,
e.g. 15 + 180 or 195 also works.
your equation is
2cos2x = 1
cos2x = 1/2
2x = 30 degrees or 2x = 150 degrees
x = 15 or x = 75 degrees
you state no domain, but since the period of cos2x is 180 degrees, more answers can be obtained by simply adding/subtracting multiple of 180 to the above answers,
e.g. 15 + 180 or 195 also works.
go with drwls solution, I was thinking sine
mine is wrong at the end
mine is wrong at the end
cosx+ sqrt2*sinx= sqrt3
Divide both sides by sqrt3
(1/sqrt3)*cosx + sqrt(2/3)*sinx = 1
Note that 1/sqrt3 = cos 54.74 degrees
and sqrt(2/3) = sin 54.74 degrees
Now use the "transformation" identity
cosa cosb + sina sinb = cos(a-b)
cos(a - 54.74deg)= 1
a -54.74 = 0 degrees
a = 54.74 deg
Check:
cosx = 0.5773
sqrt2*sinx = 1.1548
sum = 1.732 looks good
Divide both sides by sqrt3
(1/sqrt3)*cosx + sqrt(2/3)*sinx = 1
Note that 1/sqrt3 = cos 54.74 degrees
and sqrt(2/3) = sin 54.74 degrees
Now use the "transformation" identity
cosa cosb + sina sinb = cos(a-b)
cos(a - 54.74deg)= 1
a -54.74 = 0 degrees
a = 54.74 deg
Check:
cosx = 0.5773
sqrt2*sinx = 1.1548
sum = 1.732 looks good
the second: ...
cosx + √2sinx = √3
√2sinx = √3 - cosx
square both sides
2sin^2 x = 3 - 2√3cosx + cos^2x
2(1 - cos^2x) = 3 - 2√3cosx + cos^2x
which simplifies to
3cos^2 x - 2√3cosx + 1 = 0
solving this quadratic for cosx yields
cos x = (2√3 ± √0)/6 = √3/3
x = 54.7356 degrees or 305.2644 degrees
since I squared, all answers have to be verified.
subbing back in, x = 54.7356 works, but x = 305.2644 does not work, so
x = 54.7356 degrees
I also did this using Calculus,
consider y = cosx + √2sinx
dy/dx = -sinx + √2cosx
= 0 for a max/min
then sinx = √2cosx
sinx/cosx = √2
tanx = √2
x = arctan √2 = 54.7356 degrees
subbing this back gave me 1.7321.. which is √3
cosx + √2sinx = √3
√2sinx = √3 - cosx
square both sides
2sin^2 x = 3 - 2√3cosx + cos^2x
2(1 - cos^2x) = 3 - 2√3cosx + cos^2x
which simplifies to
3cos^2 x - 2√3cosx + 1 = 0
solving this quadratic for cosx yields
cos x = (2√3 ± √0)/6 = √3/3
x = 54.7356 degrees or 305.2644 degrees
since I squared, all answers have to be verified.
subbing back in, x = 54.7356 works, but x = 305.2644 does not work, so
x = 54.7356 degrees
I also did this using Calculus,
consider y = cosx + √2sinx
dy/dx = -sinx + √2cosx
= 0 for a max/min
then sinx = √2cosx
sinx/cosx = √2
tanx = √2
x = arctan √2 = 54.7356 degrees
subbing this back gave me 1.7321.. which is √3
What a way to spend Sunday morning, lol
... and on top of all I didn't even use the method of transformation that was asked for...
Oh well, some days .....
... and on top of all I didn't even use the method of transformation that was asked for...
Oh well, some days .....
I'm glad we agreed.
I figured out the 'scale factor' needed change the coefficients on the left, to enable a "transformation", while scribbling on the back of a breakfast napkin.
I figured out the 'scale factor' needed change the coefficients on the left, to enable a "transformation", while scribbling on the back of a breakfast napkin.
I was thinking to use the fact that any linear combination such as
√2sinx + cosx can be written as
k(sin(x+T) , where T is the translation and k is the net amplitude.
k(sinxcosT + cosxsinT) = √2sinx + cosx
so
ksinxcosT = √2sinx and kcosxsinT= cosx
cosT = √2/k and sinT = 1/k
then tanT = 1/√2
T = 35.26 degrees, the angle in a 1: √2:3 triangle
then back in sinT = 1/k, k= √3
so k(sin(x+T) = √2sinx + cosx = √3
becomes
√3sin(x + 35.264) = √3
sin(x + 35.264) = 1
x + 35.264 = 90
x = 54.74
How is that for "overkill"
I liked the way you recognized the 1 : √2 : 3 ratio right away.
√2sinx + cosx can be written as
k(sin(x+T) , where T is the translation and k is the net amplitude.
k(sinxcosT + cosxsinT) = √2sinx + cosx
so
ksinxcosT = √2sinx and kcosxsinT= cosx
cosT = √2/k and sinT = 1/k
then tanT = 1/√2
T = 35.26 degrees, the angle in a 1: √2:3 triangle
then back in sinT = 1/k, k= √3
so k(sin(x+T) = √2sinx + cosx = √3
becomes
√3sin(x + 35.264) = √3
sin(x + 35.264) = 1
x + 35.264 = 90
x = 54.74
How is that for "overkill"
I liked the way you recognized the 1 : √2 : 3 ratio right away.
1024*(GT430)