If I had the equation 2cosx + sinx = 1 then I could express it as
√5 cos (x-26.6) = 1. I have no problem when it's like this :)
However, if it was 2sinx + cosx = 1, how could I do it? would it be
√5 sin (x - 63.4) = 1?
If I go further to find the values of x here, I calculate
sin^(-1) of 1/√5 which is 26.6, and then for x I get 90 and 396.8, but because it's not under 360 I take that value away and get 36.8.
So, I said x was 36.8 and 90. Whenever I put these back into the equation though, it doesn't result in 1, it results in two. Does anyone know why this is and can point out where I'm going wrong?
Thanks if you can help :)
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Verified answer
2sin(x) + cos(x) = 1
a*sin(x + b) = a*sin(x)cos(b) + a*cos(x)sin(b) = 1
We want a*cos(b) = 2 and a*sin(b) = 1
Then by the fact that sin^2(x) + cos^2(x) = 1, we find a = sqrt(5)
Then we have cos(b) = 2/sqrt(5) and sin(b) = 1/sqrt(5)
Then b = 0.4636476090008061162 radians or 26.565051177077989350 deg
Then we have sqrt(5)*sin(x+b) = 1
or sin(x+b) = 1/sqrt(5)
x + b = arcsin(1/sqrt(5)) = b
x = 0 radians or 0 degrees
This makes sense since
2*sin(0) + cos(0) = 2*0 + 1 = 1