Question

Solve to find \[A\]: \[2\sin A\cos A - \cos A - 2\sin A + 1 = 0\]

Answer 1

Hint: Here, in the given question, we are given a trigonometric equation and asked to find the value of \[A\].
To solve for the value of \[A\] in the equation given, at first we will check if there is any possibility of simplifying the equation directly or if there is a need to apply any trigonometric identity. Check and continue to simplify the equation till possible extent. At last, we will get our desired answer. We can also check our answer by putting the final answer back to the equation.

Complete step-by-step solution:
Given problem,
\[2\sin A\cos A - \cos A - 2\sin A + 1 = 0\]
Rearranging the equation, we get,
\[2\sin A\cos A - 2\sin A - \cos A + 1 = 0\]
Take \[2\sin A\] common from first two terms and \[ - 1\] from the last two terms,
\[2\sin A\left( {\cos A - 1} \right) - 1\left( {\cos A - 1} \right) = 0\]
Now, again taking common\[\left( {\cos A - 1} \right)\]both sides, we get,
\[\left( {2\sin A - 1} \right)\left( {\cos A - 1} \right) = 0\] \[\]
Product of two terms is zero in the above equation. It can happen only and only when at-least one of the two numbers is zero. It means,
Either\[\left( {2\sin A - 1} \right) = 0\] or \[\left( {\cos A - 1} \right) = 0\]
If \[\left( {2\sin A - 1} \right) = 0\], then,
\[
  2\sin A = 1 \\
   \Rightarrow \sin A = \dfrac{1}{2} \\
   \Rightarrow \sin A = \sin 30^\circ \\
   \Rightarrow A = 30^\circ \\
 \]
If\[\left( {\cos A - 1} \right) = 0\], then,
\[
  \cos A = 1 \\
   \Rightarrow \cos A = \cos 0^\circ \\
   \Rightarrow A = 0^\circ \\
 \]
Hence the solution is \[A = 0^\circ \] or/and \[A = 30^\circ \]

Note: Whenever we face such types of questions, at first always check if rearrangement of the equation is beneficial for continuing to solve, and then check if any trigonometric identity can be used or not, if it is applicable then apply and continue to solve. We should remember all the basic trigonometric identities. We should just try not to complicate but simplify the equation.
As, here in the above equation we have rearranged the equation first according to the requirement and then taken factors common wherever possible and at last we got the desired answer.