Find all solutions of the equation in the interval[0, 2π) 10cosx−5√2=0

Answer 1

Another way to approach this question without a calculator:

10 cosx - 5√2 = 0

Rearrange and solve for cosx

cosx = √2 / 2 (multiple top and bottom by √2/√2 and simply to get...)

cosx = 1 / √2 (you should recognize this as a familiar form!)

Remember that cosine is adjacent / hypotenuse (SOH-CAH-TOA).

The triangle with the adjacent side of length 1 and hypotenuse of √2 is a right triangle with angles 45-45-90.

So, we know one solution to this problem is x = 45 degrees or π/4 radians.

From the interval [0, 2π), we have a full circle. Are there any other solutions?

Well, we know that cosine is positive in quadrants 1 and 4. So, a triangle in quadrant 4 would be at 7π / 4 radians.

Answer 2

First solve for the cos x

10cosx - 5√2 = 0

10cosx = 5√2

cos x = √2/2

Solving for x gives x = π/4 as one solution

Because cos x is positive, the other solution lies in the fourth quadrant

Answer 3

10cosx= 5/sqr2, divide both sides by 10

cosx =sqr2/2

use a calculator with an inverse trig function to get

x = cos^-1(.707) = 45 degrees or pi/4

and also 360-45 = 315 degrees or 7pi/4

cosines are only positive in quadrants I and IV