Verified answer

Using identities here can be very messy, so we could approach it from a graphing approach.

Cos(3x )= cos (0.5x), for 0 ≤ x ≤ π

Cos(3x) = cos(x/2)

If 0 ≤ x ≤ π, then 0 ≤ x/2 ≤ π/2 , and 0 ≤ 3x ≤ 3π

So we know that x/2 is an angle in quadrant I, and angle 3x must be in quadrant I or IV.

For x/2 and 3x to have the same cosine,

either 3x= x/2 or 3x = 2pi-x/2, or 3x = x/2 + 2pi

X= 0; 4pi/7 , or 4pi/5

You can verify these answers by graphing and finding the zeroes: y= cos(3x)- cos(x/2)

I hope this helps!

cos 3x =cos 0.5x; since outside the cos we got nothing we could safely say that all the solutions are found within:

3x = 0.5x

this gives:

6x = 0 or

x = 0

which is found within 0 ≤ x ≤ π

3x = ± 0.5x + 2nπ...for n = 0, 1, 2,..

so, 5x/2 = 2nπ or 7x/2 = 2nπ

i.e. x = 4nπ/5 or x = 4nπ/7

=> x = 0, 4π/7 and 4π/5...for 0 ≤ x ≤ π

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