Question

Find all the values of $ x $ , if the trigonometric equation $ \sin \left( {2x} \right) = 4\cos \left( x \right) $ holds.

Answer 1

Hint: First of all let’s look at the given trigonometric equation. And then we use some suitable formulae we know and solve for the value of $ x $ . We finally arrive at a simple trigonometric equation. From that equation, we can easily determine the value of $ x $ .

Complete step-by-step answer:
First of all, let’s look at the given trigonometric equation.
 $ \sin \left( {2x} \right) = 4\cos \left( x \right) $ ,
Let’s use the suitable formulae, the suitable formula is
 $ \sin \left( {2x} \right) = 2\sin x\cos x $
Let’s apply the above formula in the given trigonometric equation, we get
 $ \sin \left( {2x} \right) = 2\sin x\cos x = 4\cos x $
 $ \Rightarrow 2\sin x\cos x = 4\cos x $ ,
Let’s divide the whole equation by 2, we get
 $ \Rightarrow \sin x\cos x = 2\cos x $ ,
Let’s take all the expressions into one side of the equation.
 $ \Rightarrow \sin x\cos x - 2\cos x = 0 $ ,
Let's use the $ \cos x $ term in both expressions.
 $ \Rightarrow \cos x\left( {\sin x - 2} \right) = 0 $ ,
We know that, if $ ab = 0 \Rightarrow a = 0 $ or $ b = 0 $ , so by applying this algorithm to the above equation, we get
 $ \Rightarrow \cos x = 0 $ or $ \sin x - 2 = 0 $ ,
On further simplifications, we get
 $ \Rightarrow \cos x = 0 $ or $ \sin x = 2 $
But we know that the value of $ \sin x $ lies between $ - 1 $ and $ 1 $ .
Therefore there is no $ x $ such that $ \sin x = 2 $
Now we left only with the equation, that is
 $ \cos x = 0 $ ,
We have the values of x when $ \cos x = 0 $ is $ x = 2n\pi \pm \dfrac{\pi }{2} $ .
So, The values of $ x $ that satisfy the trigonometric equation $ \sin \left( {2x} \right) = 4\cos \left( x \right) $ are the same as the equation $ \cos x = 0 $ , that is $ x = 2n\pi \pm \dfrac{\pi }{2} $ .

Note: Observe the given trigonometric equation $ \sin \left( {2x} \right) = 4\cos \left( x \right) $ everyone tries to cancel out the term $ \cos x $ after some simplification. But we should not cancel them when it is equal to $ 0 $ . Since we cancel the $ \cos x $ term we will leave out only with the term $ \sin x = 2 $ which has no solutions. So, we must be careful while canceling the terms not only in the trigonometry, we should also check the term whether it can be equal to $ 0 $ or not while we are going to cancel it.