Me again: cos(2x)cos(π/3) + sin(2x)sin(π/3) = √3/2 for [0, 2π]?

[ahat]

1/2 cos(2x) + √3/2 sin(2x) = √3/2
cos(2x) + √3 sin(2x) = √3
cos(2x) + 2√3[sin(x)cos(x)] = √3

then I'm not sure

[Alwin]

1/2 cos(2x) + √3/2 sin(2x) = √3/2
cos(2x) + √3 sin(2x) = √3
cos(2x) + 2√3[sin(x)cos(x)] = √3

then I'm not sure

Perhaps try the identity
then solve the quadratic

[availn]

Isn't this just an addition formula? cos(a-b) = sin(a)sin(b) + cos(a)cos(b)

[ahat]

I'm sorry, I still can't do it Can I have another hint?

[Lasercookie]

Isn't this just an addition formula? cos(a-b) = sin(a)sin(b) + cos(a)cos(b)

Take a close look at the right side of the addition formula, and the left side of your equation.

Once you figure that out, spoiler for the next part if you need it:

Spoiler

Trig functions of the form are easy enough to solve. If it helps to think of it like this, let then solve .

Spoiler

once you know what u is, sub it back into u = a - b and solve for x

[lzxnl]

If you're REALLY desperate...
make the sub t= tan x
sin 2x = (2t/(1+t^2)
cos 2x = (1-t^2)/(1+t^2)
tan 2x = 2t/(1-t^2)

Convert to a quadratic equation and then solve. This substitution is useful and it'll always reduce trig expressions to polynomials, but it can get messy.

Alternatively, you have 1/2 cos 2x + sqrt3 /2 sin 2x = sqrt3 /2
Note that sin pi/6 = 1/2 and cos pi/6 = sqrt3 /2
So you really have sin pi/6*cos 2x + cos pi/6*sin 2x on the right hand side.
Does that look like anything?

[Dayman]

I think it's been said twice already use cos(x-y) then solve.

Spoiler

in this case x=2x and y=pi/3

[Alwin]

1/2 cos(2x) + √3/2 sin(2x) = √3/2
cos(2x) + √3 sin(2x) = √3
cos(2x) + 2√3[sin(x)cos(x)] = √3

then I'm not sure

I really dont mean to treat you like an idiot (sorry) but I'll write it out just one last time. Many substitution methods are possible, but most direct way:




I'm guessing that you can get it from here, otherwise check out the spoiler:

Spoiler

[ahat]

I'm sorry, I still can't do it Can I have another hint?

Wow man, hahahahaha, why did I not read this. So simple argh! I'm killing myself. Thank you so much everyone, and I'm basking in the awe of your beast scores.

Finally (ffs): pi/12, pi/4, 13pi/12, 5pi/4

[ahat]

Damn, I mean to quote avalin, "Isn't this just an addition formula? cos(a-b) = sin(a)sin(b) + cos(a)cos(b)"