
How do you solve $ {\sin ^2}x{\cos ^5}x = \left( {{{\sin }^2}x - 2{{\sin }^4}x + {{\sin }^6}x} \right)\cos x $ ?
Answer
530.7k+ views
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.
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.
Recently Updated Pages
Master Class 11 Business Studies: Engaging Questions & Answers for Success

Master Class 11 Economics: Engaging Questions & Answers for Success

Master Class 11 Computer Science: Engaging Questions & Answers for Success

Master Class 11 English: Engaging Questions & Answers for Success

Master Class 11 Maths: Engaging Questions & Answers for Success

Master Class 11 Biology: Engaging Questions & Answers for Success

Trending doubts
One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

Discuss the various forms of bacteria class 11 biology CBSE

Draw a diagram of a plant cell and label at least eight class 11 biology CBSE

State the laws of reflection of light

Explain zero factorial class 11 maths CBSE

10 examples of friction in our daily life

