Question

Solve: $ {{\sin }^{2}}{{60}^{\circ }}+{{\cos }^{2}}\left( 3x-{{9}^{\circ }} \right)=1 $ .

Answer 1

Hint: At first take the term $ {{\cos }^{2}}\left( 3x-{{9}^{\circ }} \right) $ from left to right hand side so we get, $ {{\sin }^{2}}{{60}^{\circ }}=1-{{\cos }^{2}}\left( 3x-{{9}^{\circ }} \right) $ after that we will use the identity $ {{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1 $ and use it as $ 1-{{\cos }^{2}}\theta ={{\sin }^{2}}\theta $ and then take $ \theta $ as $ 3x-{{9}^{\circ }} $ and then further solve it to find value of x.

Complete step-by-step answer:
In the question we are given the equation $ {{\sin }^{2}}{{60}^{\circ }}+{{\cos }^{2}}\left( 3x-{{9}^{\circ }} \right)=1 $ and we have to find the value of x such that it satisfies the equation.
So, we are given the equation, $ {{\sin }^{2}}{{60}^{\circ }}+{{\cos }^{2}}\left( 3x-{{9}^{\circ }} \right)=1 $ .
Now we will take $ {{\cos }^{2}}\left( 3x-{{9}^{\circ }} \right) $ left to right hand side so we get,
 $ {{\sin }^{2}}{{60}^{\circ }}=1-{{\cos }^{2}}\left( 3x-{{9}^{\circ }} \right) $ .
We all know the identity that $ {{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1 $ .
So, we can also write it as $ {{\sin }^{2}}\theta =1-{{\cos }^{2}}\theta $ now will take $ \theta $ as $ 3x-{{9}^{\circ }} $ so we get,
 $ {{\sin }^{2}}{{60}^{\circ }}=1-{{\cos }^{2}}\left( 3x-{{9}^{\circ }} \right) $ or, $ {{\sin }^{2}}{{60}^{\circ }}={{\sin }^{2}}\left( 3x-{{9}^{\circ }} \right) $ .
So from this we can write that $ {{60}^{\circ }}=3x-{{9}^{\circ }} $ .
Now subtracting $ 3x $ from both the sides we get,
 $ {{60}^{\circ }}-3x=3x-{{9}^{\circ }}-3x $ or, $ {{60}^{\circ }}-3x=-{{9}^{\circ }} $ .
Now subtracting $ {{60}^{\circ }} $ from both the sides we get,
 $ {{60}^{\circ }}-3x-{{60}^{\circ }}=-{{9}^{\circ }}-{{60}^{\circ }} $ or, $ -3x=-69 $ .
Hence the value of x will be $ \dfrac{-69}{-3} $ or, 23
So, the answer is 23.


Note: Also we can do this question by another way by directly using the identity that $ {{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1 $ and compare it with $ {{\sin }^{2}}{{60}^{\circ }}+{{\cos }^{2}}\left( 3x-{{9}^{\circ }} \right)=1 $ . Then we can say that $ 3x-{{9}^{\circ }}={{60}^{\circ }} $ and then solve it to get the value of x.