i need help on this trigonometry/algebra question

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Solve for x

f(x) = cox x - 1/2(cos 2x) = 0

So first I subsituted (2 cos^2 X - 1) for (cos 2x) using the double angle theorem.

now we have cos x - 1/2 (2 cos^2 X -1) = 0

then we distribute the 1/2

now we have cos x - (cos^X - .5)

simplify cos x - cos^2 X + .5

So now what do I do? If you divide each part by cos x, then youll have .5 over cos x, so how do isolate cosx so I can solve for x?

Substitute u for cos x.

You will then have a quadratic equation that you can solve for u, using the quadratic formula

All x for which cos x = one of the solutions for u are therefore solutions to your original equation.

the secant is cosine ^-1 or 1/cosine

that is if one of the angles is a right angle.

You're steps look right so far. Now what you have is a quadratic equation in cos x. Substitute u for cos x, you have,

u - u^2 + 1/2 = 0 or

u^2 - u - 1/2 = 0.

Solve by the quadratic formula:

u = (-b+-sqrt(b^2 - 4ac))/2a, where I've used sqrt for square root. So then

u = (1+-sqrt(3))/2.

Substituting the cos x back in, we get

cos x = (1+sqrt(3))/2 or cos x = (1-sqrt(3))/2.

Since (1+sqrt(3))/2 > 1, cos x = (1+sqrt(3))/2 has no solutions. The other equation is fine though, so the arc cosine of (1-sqrt(3))/2 would give you the final solution :

x = 111.470 degrees. I can also give to in radians if you like.

don't forget that when v is a solution - v is also a solution, as we are talking about cosines. Also if you add k*2*Pi it will still be a solution (assuming k is an integer).