# Math Help - Trigonometric Equation

1. ## Trigonometric Equation

(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 $\sqrt{13}\ cos\ (x - 33.7)$

and in (b), I got up to saying $cos\ (x - 33.7) = \frac{1}{\sqrt1{3}}$, 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

2. Originally 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 $\sqrt{13}\ cos\ (x - 33.7)$

and in (b), I got up to saying $cos\ (x - 33.7) = \frac{1}{\sqrt1{3}}$, 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 relationship $\sin{x}=\sqrt{1-\cos^2{x}}$.

So we have

$3\cos{x}+2\cdot\sqrt{1-\cos^2{x}}=1$

Square both sides

$(3\cos{x}-1)^2=(-2\cdot\sqrt{1-\cos^2{x}})^2$

$9\cos^2{x}-6\cos{x}+1=4-4\cos^2{x}$

Set the equation equal to zero and solve the quadratic in x.

$13\cos^2{x}-6\cos{x}-3=$

$\cos{x}=\frac{-(-6)\pm\sqrt{(-6)^2-4(13)(-3)}}{2(13)}$

Therefor $x=\cos^{-1}\left[\frac{-(-6)\pm\sqrt{(-6)^2-4(13)(-3)}}{2(13)}\right]$

3. Originally Posted by VonNemo19
For b) Put the equation in terms of cosine (or sine) by making use of the pythagorean relationship $\sin{x}=\sqrt{1-\cos^2{x}}$.

So we have

$3\cos{x}+2\cdot\sqrt{1-\cos^2{x}}=1$

Square both sides

$(3\cos{x}-1)^2=(-2\cdot\sqrt{1-\cos^2{x}})^2$

$9\cos^2{x}-6\cos{x}+1=4-4\cos^2{x}$

Set the equation equal to zero and solve the quadratic in x.

$13\cos^2{x}-6\cos{x}-3=$

$\cos{x}=\frac{-(-6)\pm\sqrt{(-6)^2-4(13)(-3)}}{2(13)}$
Therefor $x=\cos^{-1}\left[\frac{-(-6)\pm\sqrt{(-6)^2-4(13)(-3)}}{2(13)}\right]$