
simple trig equation
I am stuck on this trig equation.
Solve the Equation for one function of $\displaystyle \theta$ and find the angle.
$\displaystyle sin\theta + 3 cos\theta = 2 \sqrt{2}$
ok, step one is write in terms of one function. so looking at it I can't see any easy identity to replace. so I raise power on both sides.
$\displaystyle
\begin{aligned}
sin\theta + 3 cos\theta = 2 \sqrt{2}&=(sin\theta + 3 cos\theta)^2 = (2\sqrt{2})^2 \\
&=sin^2\theta + 6cos\theta sin\theta+9cos^2\theta = 8 \\
&=sin^2\theta + 6cos\theta sin\theta+9(1sin^2\theta) = 8 \\
&=8sin^2\theta  6cos\theta sin\theta = 1
\end{aligned}
$
if I raise power again it will not square the other cos. and I can't see any elementary identity to use on it. in the book it says the answer should be $\displaystyle \frac{\pi}{4}, 45^{\circ}$


Quote:
Originally Posted by
skeeter
If you wish to avoid an error click this link: http://www.mathhelpforum.com/mathhe...a172498.html (Wink)

hmm, I took a look but they say to do this problem with no calculator and just the elementary identities. (Thinking)

so I found a way to do this one and get correct solution with just the pencil. but I introduce extraneous solutions so I have to limit the domain.
$\displaystyle
\begin{aligned}
sin\theta +3cos \theta=2\sqrt{2}
&\equiv (3\sqrt{1sin^2\theta})^2 = (2\sqrt{2}sin\theta)^2 \\
&\equiv 10sin^2\theta4\sqrt{2}sin\theta1= 0 \\
&\equiv (10sin^2\theta1)^2= (4\sqrt{2}sin\theta)^2 \\
&\equiv 100sin^4\theta52sin^2\theta+1= 0 \\
&\equiv (50sin^2\theta1)(2sin^2\theta1) \\
&\equiv \bigg( sin\theta=\frac{1}{\sqrt{50}} \bigg) \bigg( sin\theta=\frac{1}{\sqrt{2}} \bigg) \\
& \therefore \theta=\frac{\pi}{4},\frac{3\pi}{4}, \lbrace 0<\theta<2\pi \rbrace \\
\end{aligned}.
$

Since the right hand side is positive and greater than 1, sinθ and cosθ must be in the firs quadrant.