Asked by Anonymus
Solve: log(2 cos(2x)+11)=1 where x e(0, 2pi)
Answers
Answered by
oobleck
log(2 cos(2x)+11)=1
now recall that 10^(logx) = x. So,
2 cos(2x) + 11 = 10^1
2 cos(2x) = -1
cos(2x) = -1/2
So, what do you think?
now recall that 10^(logx) = x. So,
2 cos(2x) + 11 = 10^1
2 cos(2x) = -1
cos(2x) = -1/2
So, what do you think?
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