find solutions in the interval [0, 2pi)
cos 2x + cos x = 0
cos 2x + cos x = 0
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I am interpreting this problem as:
cos(2*x) + cos(x) = 0
cos(2x) = cos^2(x) - sin^2(x) is a basic trig identity. Now we have:
(Equation 1) cos^2(x) + cos(x) - sin^2(x) = 0
Another well-known trig identity is cos^2(x) + sin^2(x) = 1, so when
we subtract cos^2(x) from both sides of that equation we get:
sin^2(x) = (1 - cos^2(x)). Substitute this into (Equation 1)
cos^2(x) + cos(x) - (1 - cos^2(x)) = 0
cos^2(x) + cos(x) - 1 - (-cos^2(x)) = 0 (and since minus a minus is plus)
2cos^2(x) + cos(x) - 1 = 0
This is now nothing but a simple quadratic equation with the variable being cos(x).
To make it more obvious, let u = cos(x). Then:
2u^2 + u - 1 = 0
(2u - 1) * (u + 1) = 0
So u = (1/2) and (-1) and since u = cos(x)
cos(2*x) + cos(x) = 0
cos(2x) = cos^2(x) - sin^2(x) is a basic trig identity. Now we have:
(Equation 1) cos^2(x) + cos(x) - sin^2(x) = 0
Another well-known trig identity is cos^2(x) + sin^2(x) = 1, so when
we subtract cos^2(x) from both sides of that equation we get:
sin^2(x) = (1 - cos^2(x)). Substitute this into (Equation 1)
cos^2(x) + cos(x) - (1 - cos^2(x)) = 0
cos^2(x) + cos(x) - 1 - (-cos^2(x)) = 0 (and since minus a minus is plus)
2cos^2(x) + cos(x) - 1 = 0
This is now nothing but a simple quadratic equation with the variable being cos(x).
To make it more obvious, let u = cos(x). Then:
2u^2 + u - 1 = 0
(2u - 1) * (u + 1) = 0
So u = (1/2) and (-1) and since u = cos(x)