# simple trig equation

• Feb 25th 2011, 12:23 PM
skoker
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(1-sin^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 $\displaystyle \frac{\pi}{4}, 45^{\circ}$
• Feb 25th 2011, 12:36 PM
skeeter
• Feb 25th 2011, 12:55 PM
e^(i*pi)
Quote:

Originally Posted by skeeter

If you wish to avoid an error click this link: http://www.mathhelpforum.com/math-he...-a-172498.html (Wink)
• Feb 25th 2011, 01:17 PM
skoker
hmm, I took a look but they say to do this problem with no calculator and just the elementary identities. (Thinking)
• Feb 25th 2011, 05:32 PM
skoker
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{1-sin^2\theta})^2 = (2\sqrt{2}-sin\theta)^2 \\ &\equiv 10sin^2\theta-4\sqrt{2}sin\theta-1= 0 \\ &\equiv (10sin^2\theta-1)^2= (4\sqrt{2}sin\theta)^2 \\ &\equiv 100sin^4\theta-52sin^2\theta+1= 0 \\ &\equiv (50sin^2\theta-1)(2sin^2\theta-1) \\ &\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}.
• Feb 25th 2011, 11:13 PM
sa-ri-ga-ma
Since the right hand side is positive and greater than 1, sinθ and cosθ must be in the firs quadrant.