Trigonometric Equation

• Jul 6th 2009, 12:01 PM
db5vry
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 $\displaystyle \sqrt{13}\ cos\ (x - 33.7)$

and in (b), I got up to saying $\displaystyle cos\ (x - 33.7) = \frac{1}{\sqrt1{3}}$, 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 (Nod)
• Jul 6th 2009, 12:21 PM
VonNemo19
Quote:

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

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

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

So we have

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

Square both sides

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

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

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

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

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

Therefor $\displaystyle x=\cos^{-1}\left[\frac{-(-6)\pm\sqrt{(-6)^2-4(13)(-3)}}{2(13)}\right]$
• Jul 6th 2009, 01:37 PM
db5vry
Quote:

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

So we have

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

Square both sides

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

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

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

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

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