(sqrt2)cosx + sinx = 1?

Question

solve for between 0 using either the 't' formula
or
the acosx + bsinx = Rcos (x-y), where R = sqrt(a^2 + b^2) and tany = b/a
I find neither works HELP!

Answer 1

Using the 't' method, where t = tan(x/2):
[ (sqrt(2)) (1 - t^2) + 2t - 1 - t^2 ] / [ 1 + t^2 ] = 0
-( 1 + sqrt(2) ) t^2 + 2t + (sqrt(2) - 1) = 0
(1 + sqrt(2) ) t^2 - 2t - ( sqrt(2) - 1 ) = 0
t = ( 2 +/- sqrt(4 + 4 (sqrt(2) - 1)(sqrt(2) + 1) ) / 2( 1 + sqrt(2) )
= ( 2 +/- 2sqrt(2) ) / 2( 1 + sqrt(2) )
= ( 1 +/- sqrt(2) ) / (1 + sqrt(2) )

Either:
t = ( 1 + sqrt(2) ) / ( 1 + sqrt(2) )
t = 1
x/2 = 45deg as 0 < x/2 < 180deg
x = 90deg

or

t = ( 1 - sqrt(2) ) / ( 1 + sqrt(2) ).
Multiplying numerator and denominator by 1 - sqrt(2), this gives:
t = - 1 + 2 - 2sqrt(2) ) / (1 - 2)
= 2sqrt(2) - 3
x/2 = 180deg + arctan(2sqrt(2) - 3) = 180 - 9.7356deg. as 0 < x/2 < 180deg
x = 340.53deg.

Answer 2

.sq. the two aspects so 2cosx=a million/9 .Divide the two aspects by using 2 so cosx=a million/18 . arc cos the two aspects so x=cos^-a million(a million/18). this could supply u the 1st answer. . locate all different solutions between -a hundred and eighty to a hundred and eighty tiers utilising the 'forged' approach. look it up if u don't be responsive to what that's .Use the ranges to radians formula to get them in terms of pi

Answer 3

sqrt2 cosx+sinx=1
both sides squarred
2cos^2(x)+2sqrt(2)sinx cosx+sin^2(x)=1
since sin^2(x)+cos^2(x)=1
then
2cos^2(x)+2sqrt(2)sinx cos x+1-cos^2(x)=1
cos^2(x)+2sqrt(2)sinxcosx=0
fisrt answer: cos(x)=0 so x=90 or 270
second:
cos(x)+2sqrt(2)sinx=0
cos(x)=-2sqrt(2)sinx
tanx=-1/(2sqrt(2))
then x=180-19.47=160.53 or 270+19.47=289.47
there is something wrong with the secong answer but the method is correct

Answer 4

square both sides
2(cosx)^2+(sinx)^2+2sqrt2cosxsinx=1
simplify using (cosx)^2+(sinx)^2=1
cosx=-2sqrt2sinx
tanx=1/(-2sqrt2)

Answer 5

(2^0.5)cosx+sinx=1
(1+2)^(0.5)cos(x-@)=1
3^0.5cos(x-35.3)=1
cos(x-35.3)=1/(3^0.5)
x=90