(a) Express 3cosx + 2sinx in the form Rcos(x – α), where R and α are constants with R 1 0 and
0° < α < 90°. [3]
(b) Find all values of x between 0° and 360° satisfying
3cosx + 2sinx = 1. [3]
For (a), I have
and in (b), I got up to saying, but I have no idea what is done after this to get the values of x.
How is this done? Thanks if you're able to help me
For b) Put the equation in terms of cosine (or sine) by making use of the pythagorean relationshipOriginally Posted by db5vry
(a) Express 3cosx + 2sinx in the form Rcos(x – α), where R and α are constants with R 1 0 and
0° < α < 90°. [3]
(b) Find all values of x between 0° and 360° satisfying
3cosx + 2sinx = 1. [3]
For (a), I have
and in (b), I got up to saying
How is this done? Thanks if you're able to help me.
So we have
Square both sides
Set the equation equal to zero and solve the quadratic in x.
By the quadratic formula
Therefor![x=\cos^{-1}\left[\frac{-(-6)\pm\sqrt{(-6)^2-4(13)(-3)}}{2(13)}\right]](http://latex.codecogs.com/png.latex?x=\cos^{-1}\left[\frac{-(-6)\pm\sqrt{(-6)^2-4(13)(-3)}}{2(13)}\right])
I see how you do that, and I see how it is right, but that's not the way I'm meant to do it. I need to solve it from the form Rcos(x - a) = 1. That's what I would like to learn how to do because it's troubled me in the pastFor b) Put the equation in terms of cosine (or sine) by making use of the pythagorean relationship
So we have
Square both sides
Set the equation equal to zero and solve the quadratic in x.
By the quadratic formula
Therefor
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