Question

Find the value of x(in deg) If $\sin 2x = \sin {60^ \circ }\cos {30^ \circ } - \cos {60^ \circ }\sin {30^ \circ }$.

Answer 1

Hint: The key observation in this question is to use the identity $\sin \left( {A - B} \right) = \sin A\cos B - \cos A\sin B$. As it can be seen that the given question is in the same form as the identity. Hence the values of the variables can be found out by comparing it with the identity.

Complete step-by-step answer:
The given question is,
$\sin 2x = \sin {60^ \circ }\cos {30^ \circ } - \cos {60^ \circ }\sin {30^ \circ }$
$\because {\text{ }}\sin \left( {A - B} \right) = \sin A\cos B - \cos A\sin B$
On comparing,
$A = {60^ \circ }$
$B = {30^ \circ }$
$\therefore {\text{ }}\sin {60^ \circ }\cos {30^ \circ } - \cos {60^ \circ }\sin {30^ \circ }{\text{ can be replaced by }}\sin \left( {{{60}^ \circ } - {{30}^ \circ }} \right)$
$ \Rightarrow $ $\sin 2x = \sin \left( {{{60}^ \circ } - {{30}^ \circ }} \right)$
$ \Rightarrow \sin 2x = \sin {30^ \circ }$
$\therefore {\text{ 2x = 3}}{{\text{0}}^ \circ }$
On dividing by 2 both sides,
$ \Rightarrow x = {15^ \circ }$
$\therefore {\text{ x = 1}}{{\text{5}}^ \circ }{\text{ is the solution of the given equation}}$
$\therefore {\text{ x = 1}}{{\text{5}}^ \circ }$ is the final answer.

Note: Remember that in first quadrant all trigonometry functions are positive, in second quadrant only $\sin \theta $ is positive, in third quadrant only $\tan \theta $ is positive and in fourth quadrant only $\cos \theta $ is positive. Calculations should be done carefully to avoid any mistake. After the final answer is found out it can be checked that whether it satisfies the original equation given in the question by simply substituting its value in the equation and if it does not satisfy the equation then the solution must be rechecked. In this question the value of x is in degree but it can be converted into radian by using the form ${1^ \circ } = \dfrac{\pi }{{{{180}^ \circ }}}radians$.