Question

a) By solving the equation sin2x=cos3x. Find value of sin18. Show your answer in the form of (sqrt(a)+b)/c. Where a, b, c are natural numbers.
b) Given (x^(2)-6x+9)^(x^(2)-4)=1, solve for all real values of x.
c) determine the sum

Answer 1

\[\sum_{k=1}^{\infty}\frac{ 4^{k}+3^{k-1}+2^{k-2} }{ 5^{k+1} }\]

Answer 2

\[\frac{ \sqrt{a}+b }{ c }\]

Answer 3

\[(x ^{2}-6x+9)^{x ^{2}-4}=1\]

Answer 4

@phi @hartnn

Answer 5

http://openstudy.com/study#/updates/4faf24d0e4b059b524fa3ba2

Answer 6

thanks for question 1

Answer 7

For question 2:
\[(x^2-6x+9)^{x^2-4}=1\]Consider the power of \(x^2-6x+9\) =0,
\[x^2 -4 =0\] Solve x.

Consider the base (i.e. \(x^2-6x+9\)) =1
\[x^2-6x+9 =1\] Solve x again.

Answer 8

For second question, credit to @experimentX

Answer 9

thanks you! :3

Answer 10

\[\sum_{k=1}^{\infty}\frac{ 4^{k}+3^{k-1}+2^{k-2} }{ 5^{k+1} }\]\[=\sum_{k=1}^{\infty}(\frac{ 4^{k}}{ 5^{k+1} }+\frac{3^{k-1}}{ 5^{k+1} }+\frac{2^{k-2} }{ 5^{k+1} })\]\[=\sum_{k=1}^{\infty}\frac{ 4^{k}}{ 5^{k+1} }+\sum_{k=1}^{\infty}\frac{3^{k-1}}{ 5^{k+1} }+\sum_{k=1}^{\infty}\frac{2^{k-2} }{ 5^{k+1} }\]\[=\frac{1}{5}\sum_{k=1}^{\infty}(\frac{4}{ 5})^{k} +\frac{1}{3\times 5}\sum_{k=1}^{\infty}(\frac{3}{ 5})^{k}+\frac{1}{2^2 \times 5}\sum_{k=1}^{\infty}(\frac{2}{ 5 })^{k} \]
3 G.S. sum to infinity

Answer 11

thanks you so much!!!

Answer 12

for second question: consider the base \[x ^{2}-6x+9\neq0\] right?

Answer 13

Hmm...
For the second question
\[(x^2-6x+9)^{x^2-4}=1\]\[((x-3)^2)^{x^2-4}=1\]\[(x-3)^{2(x^2-4)}=1\]\[\log(x-3)^{2(x^2-4)}=\log1\]\[2(x^2-4)\log(x-3)=0\]
So, \(2(x^2-4) =0\) or log (x-3) =0 <- Solve these two equations.

Answer 14

From this, we actually consider \[x ^{2}-6x+9 =1\] and \[x^2-4=0\]

Answer 15

@Callisto Many thanks!