+0

# (cos(3x+1))^2=0

+1
528
3

What is the general solution to this? How can I solve this algebraically?

Oct 9, 2017

#1
+27558
+1

This graph might give you a clue:

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Oct 9, 2017
#2
+99350
+1

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

$$3x+1 = \frac{\pi}{2} + k\pi \qquad \text{ where k is an integer}\\ 3x= \frac{\pi}{2} + k\pi -1\\ x= \frac{\pi}{6} + \frac{2k\pi}{6} -\frac{2}{6}\\ x= \frac{\pi-2+2k\pi}{6}$$

.
Oct 9, 2017
#3
+1

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

Oct 9, 2017