SOLUTION: Solve for X when Sin(X)+cos(X)=0.4 I know the answer, but do not know how to answer it in an exam (without trial and error looking at a sin and cosine graph) (Please explain in de

Algebra ->  Trigonometry-basics -> SOLUTION: Solve for X when Sin(X)+cos(X)=0.4 I know the answer, but do not know how to answer it in an exam (without trial and error looking at a sin and cosine graph) (Please explain in de      Log On


   



Question 1066743: Solve for X when Sin(X)+cos(X)=0.4
I know the answer, but do not know how to answer it in an exam (without trial and error looking at a sin and cosine graph) (Please explain in degrees, not radians)

Answer by KMST(5328) About Me  (Show Source):
You can put this solution on YOUR website!
sin%28x%29%2Bcos%28x%29=0.4
Let us square both sides of the equal sign.
The solutions of the resulting equation
will include all the solutions of the original equation,
and may include extra (extraneous) solutions,
but we can check at the end to eliminate the extraneous solutions.
%28sin%28x%29%2Bcos%28x%29%29%5E2=0.4%5E2
sin%5E2%28x%29%2Bcos%5E2%28x%29%2B2sin%28x%29cos%28x%29=0.16
Using the trigonometric identity sin%5E2%28x%29%2Bcos%5E2%28x%29=1 ,
we can simplify the equation above to
1%2B2sin%28x%29cos%28x%29=0.16
2sin%28x%29cos%28x%29=0.16-1
Using the trigonometric identity sin%282x%29=2sin%28x%29cos%28x%29 ,
we can simplify the equation above to
sin%282x%29=-0.84
Using the inverse function we find an approximate value for 2x :
2x=-57.14%5Eo (rounded).
There are many other angles that have sin%282x%29=-0.84 .
To begin with, for any angle theta , sin%28theta%29=sin%28180%5Eo-theta%29 .
Besides that, adding or subtracting any multiple of 360%5Eo
will give you a co-terminal angle with the same value for all its trigonometric functions.
So, 2x=180%5Eo-%28-57.14%5Eo%29=180%5Eo%2B57.14%5Eo=237.14%5Eo also has sin%282x%29=-0.84 ,
and so do all 2x angles differing from -57.14%5Eo or 237.14%5Eo by a multiple of 360%5Eo .
So, the solutions to the original equation would be among
%28-57.14%5Eo%29%2F2=-28.57%5Eo , %28360%5Eo-57.14%5Eo%29%2F2=302.86%5Eo%2F2=151.43%5Eo , 237.14%5Eo%2F2=118.57%5Eo , %28360%5Eo%2B237.14%5Eo%29%2F2=597.14%5Eo%2F2=298.57%5Eo ,
and other angles differing by a multiple of 360%5Eo .
highlight%28x=-28.57%5Eo%29 and highlight%28x=118.57%5Eo%29 look like solutions,
and they check when substituted into the original equation.
So do all the angles differing from one of those solutions by multiples of 360%5Eo .
On the other hand,
x=151.43%5Eo with 135%5Eo%3Cx%3C180%5Eo , and consequently
-cos%28x%29%3Esin%28x%29%3E0 is obviously an extraneous solution,
since it will yield sin%28x%29%2Bcos%28x%29%3C0 .
So is x=298.57%5Eo , with 270%5Eo%3Cx%3C315%5Eo and -sin%28x%29%3Ecos%28x%29%3E0 .