# Trigonometric Equation

• September 28th 2009, 07:04 AM
Trigonometric Equation
I need to solve $4\sin\theta + 5\cos\theta =6$

$4(1-\cos^2\theta)+5\cos\theta -6=0$

$-4\cos^2\theta +5\cos\theta -2=0$

$\cos\theta = \frac{-5\pm \sqrt{25-4(-2)(-4)}}{8}$

This gives me complex roots. In my book they solve the equation as follows:

$4\sin\theta=6-5\cos\theta$

$16\sin^2\theta =36-60\cos\theta +25\cos^2\theta$

$16-16\cos^2\theta = 36-60\cos\theta +25\cos^2\theta$

$0=41\cos^2\theta -60\cos\theta +20$

$\cos\theta = \frac{60 \pm\sqrt{3,600-4(41)(20)}}{2(41)}$

= $0.9499 or 0.5136$

How was my approach to the problem wrong?
• September 28th 2009, 07:13 AM
pacman
http://www.mathhelpforum.com/math-he...e683d2cc-1.gif

http://www.mathhelpforum.com/math-he...e380debb-1.gif

your identity is suspect: sin^2 A + cos^2 A = 1

sin^2 A = 1 - cos^2 A