What is the general solution to this? How can I solve this algebraically?
+118690 Thanks Alan,
I always think of the unit circle when I answer questions like this.
(cos(3x+1))^2=0
the cos of the angle is given by the x coordinate that it extends to on the unit circle.
x is zero on the y axis and this is when the angle is 90, 270, 450..... degrees (Alan's answer is in radians pi/2, 3pi/2, 5pi/2... radians )
so
\(3x+1 = \frac{\pi}{2} + k\pi \qquad \text{ where k is an integer}\\ 3x= \frac{\pi}{2} + k\pi -1\\ x= \frac{\pi}{6} + \frac{2k\pi}{6} -\frac{2}{6}\\ x= \frac{\pi-2+2k\pi}{6} \)
Solve for x:
cos^2(3 x + 1) = 0
Take the square root of both sides:
cos(3 x + 1) = 0
Take the inverse cosine of both sides:
3 x + 1 = π n + π/2 for n element Z
Subtract 1 from both sides:
3 x = -1 + π/2 + π n for n element Z
Divide both sides by 3:
x = -1/3 + π/6 + (π n)/3 for n element Z
