Pre-Calculus

Quote:

How to solve 4sinx= 2sin2xcos2x

I don't see any clever ways of attacking this so I guess brute force is what's left

note:

sin(2x) = 2sin(x)cos(x)
cos(2x) = 2cos ²(x) - 1

4sin(x)= 2sin(2x)cos(2x) ; divide both sides by 2

2sin(x) = sin(2x)cos(2x) ; substitute in the the formulas for sin and cos of 2x from above

2sin(x) = 2sin(x)cos(x)[size=150]([/size]2cos ²(x) - 1[size=150])[/size] ; divide both sides by 2sin(x)

1 = cos(x)[size=150]([/size]2cos ²(x) - 1[size=150])[/size] ; multiply out and rearrange things to get a cubic polynomial equal to 0

2cos ³(x) - cos(x) - 1 = 0 ; let u=cos(x)

2u ³ - u - 1 = 0

You have a cubic equation in u now. In general not very easy to solve. Since the constant term is 1 you should try to see if it's divisible by (u-1) and/or (u+1)

Lo and Behold
2u ³ - u - 1 = (u-1)[size=150]([/size]2u ² + 2u + 1[size=150])[/size] and
2u ³ - u - 1 = (u+1)[size=150]([/size]2u ² - 2u + 1[size=150])[/size]

so either
(u-1) = 0, or [size=150]([/size]2u ² + 2u + 1[size=150])[/size] = 0 or
(u+1) = 0, or [size=150]([/size]2u ² - 2u + 1[size=150])[/size] = 0

If you try to solve either of the quadratic factors on the right you'll find neither has real roots so u=1 or u=-1 are our only real solutions. So

u = 1 ==> cos(x) = 1 ==> x = pi*2k where k is any integer and
u = -1 ==> cos(x) = -1 ==> x = pi*(2k+1) where k is any integer

so our final solution is x = pi*k where k is some integer

This is a pretty tricky problem.

Question to op. What's your area of study currently?