Thread: please help me find the unknowns!

3.5 cos(x) + 2 cos (y) = 4
3.5 sin (x) = 2 sin (y) + 0.5

I already have the final answer which are x=36.87 and y=53.13
I need to know how they solved it.

thanks!!

oh and i've tried everything. adding them,dividing them,multiplying with cos(x),sin(x)..etc everything!;/

Originally Posted by maa

3.5 cos(x) + 2 cos (y) = 4
3.5 sin (x) = 2 sin (y) + 0.5

I already have the final answer which are x=36.87 and y=53.13
I need to know how they solved it.

thanks!!

oh and i've tried everything. adding them,dividing them,multiplying with cos(x),sin(x)..etc everything!;/

Your problem essentially boils down to solving the following four equations simultaneously:

3.5a + 2b = 4 => 7a + 4b = 8 .... (1)

3.5c - 2d = 0.5 => 7c - 4d = 1 .... (2)

a^2 + c^2 = 1 .... (3)

b^2 + d^2 = 1 .... (4)

where all of the a, b, c and d lie between -1 and 1 inclusive.

ok I understand how you got the 4 equations but how can i solve it? using a matrix?

7 4 0 0 8
0 0 7 -4 1
? 0 ? 0 1
0 ? 0 ? 1

is that right? if yes, what should i put in ( ? ) cause the a,b,c,d are square in this case. (is it square root to 1 which equals one?)
thanks again!

Originally Posted by maa

ok I understand how you got the 4 equations but how can i solve it? using a matrix?

7 4 0 0 8
0 0 7 -4 1
? 0 ? 0 1
0 ? 0 ? 1

is that right? if yes, what should i put in ( ? ) cause the a,b,c,d are square in this case. (is it square root to 1 which equals one?)
thanks again!

Marices can only be used to solve a system of linear equations. These equations aren't all linear. I'd substitute (1) and (2) into (3) and (4) to get two quadratic equations in a and d. Get them and solve them.