How do you solve #2cosx+3=0#?

2 Answers

#x = (2k+1)pi +- ln((3+sqrt(5))/2) i" "# for any integer #k#

Explanation:

Given:

#2 cos x + 3 = 0#

Subtracting #3# from both sides and dividing by #2# this becomes:

#cos x = -3/2#

This is outside the range #[-1, 1]# of #cos x# as a real valued function of real values. So there are no Real solutions.

What about complex solutions?

Euler's formula tells us:

#e^(ix) = cos x + i sin x#

Taking conjugate, we get :-

#e^(-ix) = cos x - i sin x#

Hence on adding above two equations we get :-

#cos x = 1/2(e^(ix) + e^(-ix))#

So the given equation becomes:

#e^(ix) + e^(-ix) + 3 = 0#

Multiplying by #4e^(ix)# and rearranging slightly:

#0 = 4(e^(ix))^2+12(e^(ix))+4#

#color(white)(0) = (2e^(ix))^2+2(2e^(ix))(3)+9-5#

#color(white)(0) = (2e^(ix)+3)^2-(sqrt(5))^2#

#color(white)(0) = (2e^(ix)+3-sqrt(5))(2e^(ix)+3+sqrt(5))#

So:

#e^(ix) = (-3+-sqrt(5))/2#

So:

#ix = ln((-3+-sqrt(5))/2) + 2kpii#

#color(white)(ix) = +-ln((3+sqrt(5))/2) + (2k+1)pii#

for any integer #k#

So:

#x = (2k+1)pi +- ln((3+sqrt(5))/2) i#

Apr 15, 2018

No solutions for #x in RR#

Explanation:

#2cosx+3=0#

Subtract #3# from both sides and divide by 2:

#cosx=-3/2#

This show that the equation has no real solutions. We know this because:

#-1<=cosx<=1#

The graph of #y=2cos(x)+3# confirms this:

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