Question

How do you find the general solution for $\sin \left( {2a} \right) - \cos \left( a \right) - 2\sin \left( a \right) + 1 = 0$?

Answer 1

If you look closely, the first term can be written in the form of an identity which further simplifies if we arrange the equation by taking out common terms. Further solving we get two solutions for the given equation which we will have to generalize to reach the final general solution.

Formula used:
The identity used in this sum is
$\sin \left( {2\theta } \right) = 2\sin \left( \theta \right)\cos \left( \theta \right)$

Complete step by step solution:
Let us try to write down the equation given.
$\sin \left( {2a} \right) - \cos \left( a \right) - 2\sin \left( a \right) + 1 = 0$
The first term of the equation can be written as $\sin \left( {2a} \right) = 2\sin \left( a \right)\cos \left( a \right)$ since we have a trigonometric identity $\sin \left( {2\theta } \right) = 2\sin \left( \theta \right)\cos \left( \theta \right)$.
We use this identity here as it will help us to get our solution easily and quickly.
Therefore we have
$2\sin \left( a \right)\cos \left( a \right) - \cos \left( a \right) - 2\sin \left( a \right) + 1 = 0$
Taking $2\sin \left( a \right)$ common from the above equation we get $2\sin \left( a \right)\left[ {\cos \left( a \right) - 1} \right] - \cos \left( a \right) + 1 = 0 \Rightarrow 2\sin \left( a \right)\left[ {\cos \left( a \right) - 1} \right] - \left[ {\cos \left( a \right) - 1} \right] = 0 \Rightarrow \left[ {2\sin \left( a \right) - 1} \right]\left[ {\cos \left( a \right) - 1} \right] = 0 \\\$
Therefore solving the above equation we get either
$\left[ {2\sin \left( a \right) - 1} \right] = 0 - - - (1)$
Or
$\left[ {\cos \left( a \right) - 1} \right] = 0 - - - (2)$
Solving equation number (1) we get
$2\sin \left( a \right) - 1 = 0 \\\ \Rightarrow 2\sin \left( a \right) = 1 \\\ \Rightarrow \sin \left( a \right) = \dfrac{1}{2} \\\$
We know that $\sin \left( a \right) = \dfrac{1}{2}$ when $a = \dfrac{\pi }{6}$ and $\dfrac{{5\pi }}{6}$.
Therefore $a = \left\{ \begin{matrix}

\dfrac{\pi }{6} + 2n\pi \\
\dfrac{{5\pi }}{6} + 2n\pi \\
\end{matrix} \right.$is the general solution when$\left[ {2\sin \left( a \right) - 1} \right] = 0$Now solving equation (2) we get$
\left[ {\cos \left( a \right) - 1} \right] = 0 \\
\Rightarrow \cos \left( a \right) = 1 \\
$We know that$\cos \left( a \right) = 1$when$a = 0$and$a = 2\pi $. Therefore$a = 2n\pi $is the general solution of$\left[ {\cos \left( a \right) - 1} \right] = 0$. Hence the general solutions for$\sin \left( {2a} \right) - \cos \left( a \right) - 2\sin \left( a \right) + 1 = 0$are$a = \left\{ \begin{matrix}
\dfrac{\pi }{6} + 2n\pi \\
\dfrac{{5\pi }}{6} + 2n\pi \\
\end{matrix} \right.$and$a = 2n\pi $ respectively.

Note: Always remember that when asked for the solution of trigonometric equations always try to find trigonometric identity which is similar to the equation given. If we can find a similar identity, we can make the required changes in the equation given. If the entire equation is not similar to any identity at least try to get some part of the equation similar to any trigonometric identity.