When is #sin(x)=\frac{24cos(x)-\sqrt{576cos^2(x)+448}}{14}#?

Question

I have this equation and I'm trying to figure out which values of x make it true:

#sin(x)=\frac{24cos(x)-\sqrt{576cos^2(x)+448}}{14}#

I checked Wolfram-Alpha and it turns out the solution is

#x=2(\pi n + arctan(2))#

for any integer n, but when I try to get that myself I have been having issues. Can anyone provide a solution or some hints?

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Answer 1 · 2017-09-15T00:57:21

#x=2pin+-sin^-1(4/5).......ninZZ#

#sin(x)=\frac{24cos(x)-\sqrt{576cos^2(x)+448}}{14}#

Rearranging we get,

#\sqrt{576cos^2(x)+448}=24cos(x)-14sin(x)#

Squaring both the sides and simplifying, we get

#16+24sin(x)cos(x)=7sin^2(x)#

#=>16+24sin(x)sqrt(1-sin^2(x))=7sin^2(x)#

#=>1-sin^2(x)=((7sin^2(x)-16)/(24sin(x)))^2#

Simplifying this further, we get the reducible quartic equation

#625sin^4(x)-800sin^2(x)+256=0#

#=>sin^2(x)=(800+-sqrt((800)^2-4*625*256))/(2*625)=16/25#

#=>color(blue)(x=2pin+-sin^-1(4/5))#