- #1

- 86

- 0

8cosx-4=0

I have come up with: x=pi/3,4pi/3

This doesn't seem correct but I am stumped.

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- Thread starter TonyC
- Start date

- #1

- 86

- 0

8cosx-4=0

I have come up with: x=pi/3,4pi/3

This doesn't seem correct but I am stumped.

- #2

VietDao29

Homework Helper

- 1,423

- 2

Remember, if you need to solve:

[tex]\cos \alpha = \cos \beta = x[/tex]

[tex]\Leftrightarrow \alpha = \pm \beta + k2\pi, k \in \mathbb{Z}[/tex]

In other words,

[tex]\alpha = \pm \arccos x + k2\pi, k \in \mathbb{Z}[/tex]

Viet Dao,

- #3

- 86

- 0

so,

5pi/3

I C

Thank you

5pi/3

I C

Thank you

- #4

HallsofIvy

Science Advisor

Homework Helper

- 41,833

- 963

TonyC said:

8cosx-4=0

I have come up with: x=pi/3,4pi/3

This doesn't seem correct but I am stumped.

8 cos x= 4 so cos x= 1/2. You could use a calculator but I think of half an equilateral triangle to observe that [itex]cos\left(\frac{\pi}{3}\right)= \frac{1}{2}[/itex]. I then recall that "cos t" is the x coordinate of the unit circle parametrized by x= cos t, y= sin t. Drawing a vertical line at x= 1/2 I notice the vertical symmetry: [itex]cos\left(-\frac{\pi}{3}\right)= \frac{1}{2}[/itex] also. That gives me two solutions, between [itex]-\pi[/itex] and [itex]\pi[/tex]. If you want solutions between 0 and [itex]2\pi[/itex], [itex]-\frac{\pi}{3}+ 2\pi= \frac{5\pi}{3}[/itex] (NOT [itex]\frac{4\pi}{3}[/itex]).

Since cosine is periodic with period [itex]2\pi[/itex],

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