Question

  cosx*cos2x*cos3x=1

Answer 1

1/4 (cos(2 x)+cos(4 x)+cos(6 x)+1) = 1

 

TrigReduce[Cos[x] Cos[2 x] Cos[3 x] == 1] 

 

(1 + Cos[2 x] + Cos[4 x] + Cos[6 x])/4 == 1 

Answer 2

We can use the trigonometric identity cos(2x) = 2cos^2(x) - 1 to rewrite cos2x as 2cos^2(x) - 1, and cos(3x) = 4cos^3(x) - 3cos(x) to rewrite cos3x as 4cos^3(x) - 3cos(x). Substituting these identities into the given equation, we have:

cos(x) * (2cos^2(x) - 1) * (4cos^3(x) - 3cos(x)) = 1

Expanding the product and simplifying, we get:

8cos^6(x) - 10cos^4(x) + 3cos^2(x) - 1 = 0

Now, let's make a substitution u = cos^2(x). Then the equation becomes:

8u^3 - 10u^2 + 3u - 1 = 0

We can use synthetic division or the rational root theorem to find that u = 1/2 is a root of this equation. This gives us the factorization:

(2u - 1)(4u^2 - 2u + 1) = 0

The quadratic factor does not have any real roots, so the only solution is u = 1/2, which means cos^2(x) = 1/2. Taking the square root of both sides, we get:

cos(x) = ±sqrt(2)/2

Therefore, the solutions to the original equation are:

x = π/4 + kπ or x = 3π/4 + kπ, where k is any integer.