Asked by Mark
Solve the equation in the given domain
cos 4x + cos 2x = 0, 0<(or equal to) x < pi
cos 4x + cos 2x = 0, 0<(or equal to) x < pi
Answers
Answered by
Reiny
use the identity
cos 2A = 2 cos^2 A - 1 on cos4A
cos 4A = 2cos^2 2A - 1
so
2cos^2 2A - 1 + cos 2A = 0
(2cos 2A -1)(cos 2A + 1) = 0
cos 2A = 1/2 or cos 2A = -1
2A = 60° or 300° or 2A = 180°
A = 30° or 150° or A = 90°
I guess I should have kept x instead of A, no big deal.
x = 30°, 90° , 150°
or
x = π/6, π/2, 5π/6
check:
if x=30° --- cos 120° + cos 60° = -1/2 + 1/2 = 0
if x=90 --- cos 360 + cos 180 = 1 - 1 =0
if x = 150 -- cos 600 + cos 300 = -1/2 + 1/2 = 0
all is good.
cos 2A = 2 cos^2 A - 1 on cos4A
cos 4A = 2cos^2 2A - 1
so
2cos^2 2A - 1 + cos 2A = 0
(2cos 2A -1)(cos 2A + 1) = 0
cos 2A = 1/2 or cos 2A = -1
2A = 60° or 300° or 2A = 180°
A = 30° or 150° or A = 90°
I guess I should have kept x instead of A, no big deal.
x = 30°, 90° , 150°
or
x = π/6, π/2, 5π/6
check:
if x=30° --- cos 120° + cos 60° = -1/2 + 1/2 = 0
if x=90 --- cos 360 + cos 180 = 1 - 1 =0
if x = 150 -- cos 600 + cos 300 = -1/2 + 1/2 = 0
all is good.
Answered by
Mark
Thanks. Makes sense. My problem was I was trying to use another double argument property.
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