How do you solve 2cosx+3=02cosx+3=0?
2 Answers
Explanation:
Given:
2 cos x + 3 = 02cosx+3=0
Subtracting
cos x = -3/2cosx=−32
This is outside the range
What about complex solutions?
Euler's formula tells us:
e^(ix) = cos x + i sin xeix=cosx+isinx
Taking conjugate, we get :-
e^(-ix) = cos x - i sin xe−ix=cosx−isinx
Hence on adding above two equations we get :-
cos x = 1/2(e^(ix) + e^(-ix))cosx=12(eix+e−ix)
So the given equation becomes:
e^(ix) + e^(-ix) + 3 = 0eix+e−ix+3=0
Multiplying by
0 = 4(e^(ix))^2+12(e^(ix))+40=4(eix)2+12(eix)+4
color(white)(0) = (2e^(ix))^2+2(2e^(ix))(3)+9-50=(2eix)2+2(2eix)(3)+9−5
color(white)(0) = (2e^(ix)+3)^2-(sqrt(5))^20=(2eix+3)2−(√5)2
color(white)(0) = (2e^(ix)+3-sqrt(5))(2e^(ix)+3+sqrt(5))0=(2eix+3−√5)(2eix+3+√5)
So:
e^(ix) = (-3+-sqrt(5))/2eix=−3±√52
So:
ix = ln((-3+-sqrt(5))/2) + 2kpiiix=ln(−3±√52)+2kπi
color(white)(ix) = +-ln((3+sqrt(5))/2) + (2k+1)piiix=±ln(3+√52)+(2k+1)πi
for any integer
So:
x = (2k+1)pi +- ln((3+sqrt(5))/2) ix=(2k+1)π±ln(3+√52)i
No solutions for
Explanation:
Subtract
This show that the equation has no real solutions. We know this because:
The graph of
