How do you solve $ \sin x - \cos x = \dfrac{1}{3} $ ?

Hint: In order to determine the solution of the above trigonometric equation , assume an angle t having value $ {45^ \circ } $ so that $ \tan t = \dfrac{{\sin t}}{{\cos t}} = 1\,\,\,as\,t = {45^ \circ } $ . Try to combine the terms on the LHS into one single trigonometric function including the assumption in the equation, and apply the formula $ \sin A\cos B - \sin B\cos A = \sin \left( {A - B} \right) $ .Since we know that the sine function if always positive in 1st and 2nd quadrant. So there will be two solutions to the given equation. For the first quadrant use the property $ {\sin ^{ - 1}}\left( {\sin x} \right) = x $ and for the 2nd quadrant use $ {\sin ^{ - 1}}\left( {\sin x} \right) = \pi - x $ property of inverse sine.

Complete step by step solution:
We are given a trigonometric equation $ \sin x - \cos x = \dfrac{1}{3} $ .
Here we will try to combine the terms in the LHS part to make it a single trigonometric function.
So, to do so let’s assume an angle $ t $ equal to $ {45^ \circ } $ . Hence the tangent of this angle $ t $ will be equal to 1. And as we know the tangent is the ratio of sine and cosine function, we can write
 $ \tan t = \dfrac{{\sin t}}{{\cos t}} = 1\,\,\,as\,t = {45^ \circ } $ ----(1)
Let's rewrite our original equation as
 $ \sin x - \left( 1 \right)\cos x = \dfrac{1}{3} $
Since from the result obtained in equation (1) we have $ 1 = \dfrac{{\sin t}}{{\cos t}} $ so replacing this in the above equation we have,
 $ \sin x - \left( {\dfrac{{\sin t}}{{\cos t}}} \right)\cos x = \dfrac{1}{3} $
Taking LCM as $ \cos t $
 $ \dfrac{{\sin x\cos t - \sin t\cos x}}{{\cos t}} = \dfrac{1}{3} $
Now multiplying $ \cos t $ on both the sides, we get
 \[
  \cos t\left( {\dfrac{{\sin x\cos t - \sin t\cos x}}{{\cos t}}} \right) = \cos t\left( {\dfrac{1}{3}} \right) \\
  \sin x\cos t - \sin t\cos x = \dfrac{{\cos t}}{3} \;
 \]
Using the formula $ \sin A\cos B - \sin B\cos A = \sin \left( {A - B} \right) $ by considering A as x and B as t
 \[
  \sin x\cos t - \sin t\cos x = \dfrac{{\cos t}}{3} \\
  \sin \left( {x - t} \right) = \dfrac{{\cos t}}{3} \;
 \]
Putting back the value of $ t = {45^ \circ } $
 \[\sin \left( {x - {{45}^ \circ }} \right) = \dfrac{{\cos {{45}^ \circ }}}{3}\]
Putting $ \cos \left( {{{45}^ \circ }} \right) = \dfrac{{\sqrt 2 }}{2} $ , we get
 \[\sin \left( {x - {{45}^ \circ }} \right) = \dfrac{{\sqrt 2 }}{6}\]
taking inverse of sine on both sides , we get
 \[{\sin ^{ - 1}}\left( {\sin \left( {x - {{45}^ \circ }} \right)} \right) = {\sin ^{ - 1}}\left( {\dfrac{{\sqrt 2 }}{6}} \right)\] ----(1)
Since $ {\sin ^{ - 1}}\left( {\sin } \right) = 1 $ as they both are inverse of each other
 \[x - {45^ \circ } = {\sin ^{ - 1}}\left( {\dfrac{{\sqrt 2 }}{6}} \right)\]
 \[
  x - {45^ \circ } = {13.59^ \circ } \\
  x = {13.59^ \circ } + 45 \;
 \]
 \[x = {58.59^ \circ }\] ---------(2)
Since the sine function is positive in 1st and 2nd quadrant both.so using the property of inverse sine function that $ {\sin ^{ - 1}}\left( {\sin x} \right) = \pi - x $ where $ x \in \left[ {\dfrac{\pi }{2},\dfrac{{3\pi }}{2}} \right] $ in the equation (1), we get
 \[
  {\sin ^{ - 1}}\left( {\sin \left( {x - {{45}^ \circ }} \right)} \right) = {\sin ^{ - 1}}\left( {\dfrac{{\sqrt 2 }}{6}} \right) \\
  x - {45^ \circ } = \pi - {\sin ^{ - 1}}\left( {\dfrac{{\sqrt 2 }}{6}} \right) \\
  x - {45^ \circ } = \pi - {13.59^ \circ } \;
 \]
The degree equivalent of $ \pi $ radian is $ {180^ \circ } $
 \[
  x - {45^ \circ } = {180^ \circ } - {13.59^ \circ } \\
  x = {166.41^ \circ } + {45^ \circ } \\
 \]
 \[x = {211.41^ \circ }\] --------(3)
From equation (2) and (3), we can conclude
 \[x = {211.41^ \circ }\,or\,x = {58.59^ \circ }\]
Therefore, the solution to the given trigonometric equation is \[x = {211.41^ \circ }\,or\,x = {58.59^ \circ }\]
So, the correct answer is “ \[x = {211.41^ \circ }\,or\,x = {58.59^ \circ }\] ”.

Note: 1.You can also convert the left-hand side of the equation using rule \[\sin x = \cos \left( {\dfrac{\pi }{2} - x} \right)\]
2. Verify your answer with the use of a calculator.
3. The equivalent degree value of $ \pi $ radian is $ {180^ \circ } $
4. $ \sin A\cos B - \sin B\cos A = \sin \left( {A - B} \right) $