How do you solve $ {\sin ^2}x{\cos ^5}x = \left( {{{\sin }^2}x - 2{{\sin }^4}x + {{\sin }^6}x} \right)\cos x $ ?

Hint: The given question involves solving a trigonometric equation and finding the value of angle x that satisfies the given equation. There can be various methods to solve a specific trigonometric equation. For solving such questions, we need to have knowledge of basic trigonometric formulae and identities.

Complete step by step solution:
In the given problem, we have to solve the trigonometric equation $ {\sin ^2}x{\cos ^5}x = \left( {{{\sin }^2}x - 2{{\sin }^4}x + {{\sin }^6}x} \right)\cos x $ and find the values of x that satisfy the given equation.
So, In order to solve the given trigonometric equation $ {\sin ^2}x{\cos ^5}x = \left( {{{\sin }^2}x - 2{{\sin }^4}x + {{\sin }^6}x} \right)\cos x $ , we should first take the terms common from the left side of the equation, we get,
 $ \Rightarrow {\sin ^2}x{\cos ^5}x = {\sin ^2}x\left( {1 - 2{{\sin }^2}x + {{\sin }^4}x} \right)\cos x $
Now, we know that the bracket $ \left( {1 - 2{{\sin }^2}x + {{\sin }^4}x} \right) $ can be condensed as $ {\left( {1 - {{\sin }^2}x} \right)^2} $ as we know that $ {\left( {a - b} \right)^2} = {a^2} - 2ab + {b^2} $ .
 $ \Rightarrow {\sin ^2}x{\cos ^5}x = {\sin ^2}x{\left( {1 - {{\sin }^2}x} \right)^2}\cos x $
Now, taking all the terms to left side of the equation, we get,
 $ \Rightarrow {\sin ^2}x{\cos ^5}x - {\sin ^2}x{\left( {1 - {{\sin }^2}x} \right)^2}\cos x = 0 $
Taking $ {\sin ^2}x\cos x $ common from the terms, we get,
 $ \Rightarrow {\sin ^2}x\cos x\left[ {{{\cos }^4}x - {{\left( {1 - {{\sin }^2}x} \right)}^2}} \right] = 0 $
Now, we know the trigonometric identity $ \left( {1 - {{\sin }^2}x} \right) = {\cos ^2}x $ . So, using the algebraic identity, we get,
 $ \Rightarrow {\sin ^2}x\cos x\left[ {{{\cos }^4}x - {{\left( {{{\cos }^2}x} \right)}^2}} \right] = 0 $
Simplifying the equation further, we get,
 $ \Rightarrow {\sin ^2}x\cos x\left[ {{{\cos }^4}x - {{\cos }^4}x} \right] = 0 $
 $ \Rightarrow {\sin ^2}x\cos x\left( 0 \right) = 0 $
So, the equation above is true for every value of x. Hence, the solution of the trigonometric equation given in the question $ {\sin ^2}x{\cos ^5}x = \left( {{{\sin }^2}x - 2{{\sin }^4}x + {{\sin }^6}x} \right)\cos x $ is all real values of the angle x.

Note: There can be various ways or methods of finding the solution set of any provided trigonometric equation. If while solving any equation, we stumble across any known fact, identity or equation, we can conclude that the solution of the given equation is all the real values of x.