Question

How do you solve $ \sqrt 3 \sin x - \cos x = 1 $ ?

Answer 1

Hint: In order to evaluate the given question , we must first know the trigonometric ratios and the trigonometric identities . Every trigonometric function and formulae are designed on the basis of three primary ratios. Sine, Cosine and tangents are these ratios in trigonometry based on Perpendicular, Hypotenuse and Base of a right triangle . In order to calculate the angles sin , cos and tan functions . For this particular question we need to know the sum or difference identity through which on applying we can get our required solution . Also we will use the concept of Equivalent equations are said to be algebraic equations that may have the same solutions if we add or subtract the same number to both sides of an equation - Left hand side or Right hand side of the equal to sign . Or we can multiply or divide the same number to both sides of an equation - Left hand side or Right hand side of the equal to sign .

Complete step-by-step answer:
For evaluating the given question , $ \sqrt 3 \sin x - \cos x = 1 $ we must recall the trigonometric identity related to this given question .
using the cosine and sine sum or difference identity fits best to the question and we can apply that having the formula as required to simplify and solve the question ,
The formula of above kind is = $ \sin (x - y) = \sin x\cos y - \cos x\sin y $ .
We will now divide the given equation , $ \sqrt 3 \sin x - \cos x = 1 $ by 2 both the sides L. H. S. and R. H. S. –
 $ \sqrt 3 \sin x - \cos x = 1 $ -----equation 1
\[\dfrac{{\sqrt 3 }}{2}\sin x - \dfrac{1}{2}\cos x = \dfrac{1}{2}\] -----equation 2
Now , we will use the cosine and sine sum or difference identity ,
The L. H. S. of the equation (2) is the same as the R. H. S. of the following trigonometric identity .
$ \sin (x - y) = \sin x\cos y - \cos x\sin y $ -----equation 3
Where $ \cos y = \dfrac{{\sqrt 3 }}{2} $ and $ \sin y = \dfrac{1}{2} $ , we know the values are possible only at angle $ \dfrac{\pi }{6} $ . Therefore , we can substitute $ \sin \left( {x - \dfrac{\pi }{6}} \right) $ at L. H. S. of the equation (2) .
 $ \sin \left( {x - \dfrac{\pi }{6}} \right) = \dfrac{1}{2} $ -----equation 4
Now , we will use the inverse sine function of both the sides of the equation (4) ,
 $ \left( {x - \dfrac{\pi }{6}} \right) = {\sin ^{ - 1}}\left( {\dfrac{1}{2}} \right) $ -------- equation 5
The sine function is always $ \dfrac{1}{2} $ at $ \theta = \dfrac{\pi }{6},\dfrac{{5\pi }}{6} $
So ,
$
  \left( {x - \dfrac{\pi }{6}} \right) = \left\{ {\dfrac{\pi }{6} + 2n\pi ,\dfrac{{5\pi }}{6} + 2n\pi } \right\} \\
  x = \left\{ {\dfrac{\pi }{3} + 2n\pi ,\pi + 2n\pi } \right\} \\
  $
Where n is any positive or negative integer including 0 .
So, the correct answer is “$ x = \left\{ {\dfrac{\pi }{3} + 2n\pi ,\pi + 2n\pi } \right\} $”.

Note: Learn the standard values of trigonometry angles by heart .
It should be very clear that \[sin\left( {A + B} \right)\]is not equal to $ \sin A + \sin B $ .
No sine or cosine can ever be more than 1 as the ratio has the hypotenuse as its denominator.
One must be careful while taking values from the trigonometric table and cross-check at least once to avoid any error in the answer.
We know that $ $ $ \sin ( - \theta ) = - \sin \theta .\cos ( - \theta ) = \cos \theta \,and\,\tan ( - \theta ) = - \tan \theta $
Therefore, a $ \sin \theta $ nd $ \tan \theta $ and their reciprocals, $ \cos ec\theta $ and $ \cot \theta $ are odd functions whereas \[\cos \theta \] and its reciprocal \[\sec \theta \] are even functions .
One must be careful while taking values from the trigonometric table and cross-check at least once to avoid any error in the answer.