How to solve $(x^2 - 11x + 29)^{(6x^2 + x - 2)}=1$? This question comes from PURE MATHEMATICS 1 for As and A levels. This question is part of exercise 1 and it has 5 Answers : $1/2, - 2/3, 4, 6 $ and $7$ . The first 2 values $1/2$ and $- 
2/3 $ I am able to find but the rest I can't. I get this first 2 values by make 1 to $(x^2 - 11x + 29)^0$ and this solving by transposition as base of power are same so they get cancelled and make it $(6x^2 + x - 2)=0.$ 

It's simple
This equation is true when either $x^2 -11x+29=1$ or $6x^2+x-2 =0$ or $x^2-11x+29=-1 \ \  and  \ \ 6x^2+x-2= even$ because $1^k=1$ and $n^0=1$ and $(-1)^{even}=0$
Solve the equations
You have solved second equation so I'll leave that
For the first  you will get $$x^2 -11x+28=0$$
$$(x-7)(x-4)=0$$ 
Solution is$ x=4,7 $as you desired
For third equation
$x^2-11x+30=0\ \ and \ \ 6x^2+x-2= even$
We get
$$x=5,6 \ \ and \ \ 6x^2+x-2=even$$
Plug in each value 
For $x=5 6x^2+x-2=153$ which is odd
But for $x=6 6x^2+x-2= 220$ which is even
So x=6 is also a solution
This method is consistent with the As and A Lvl maths
