Find the values of x which satisfy the equation 8^(1+│cos x│+ cos^2 x +│cos^3 x│+.........infinite terms) = 4^3
^ - raised to
│ │- Absolute value
Answer: {x│x = 2nπ ± π/3} U {x│x = 2nπ ± 2π/3}
How do we solve this problem?
Find the values of x which satisfy the equation 8^(1+│cos x│+ cos^2 x +│cos^3 x│+.........infinite terms) = 4^3
^ - raised to
│ │- Absolute value
Answer: {x│x = 2nπ ± π/3} U {x│x = 2nπ ± 2π/3}
How do we solve this problem?
$\displaystyle 8^{1+|\cos x| + |\cos^2 x| + |\cos^3 x| + ...}$
$\displaystyle 1+|\cos x| + |\cos^2 x| + |\cos^3 x| + ...$ is a geometric series with $\displaystyle r=|\cos x|$
so we have the sum as $\displaystyle \frac{1}{1- |\cos x|}$
and so we have $\displaystyle 8^{\frac{1}{1- |\cos x|}} = 4^3$
$\displaystyle \frac{1}{1- |\cos x|} \ln 2^3 = \ln 4^3 = \ln 2^6$
$\displaystyle 3\frac{1}{1- |\cos x|} \ln 2 = 6\ln 2$
simplifies to $\displaystyle \frac{1}{1- |\cos x|} = 2$
i hope you can finish this already..