Question

How do I solve for the two smallest positive solutions for $\sin \left( 2x \right)\cos \left( 6x \right)-\cos \left( 2x \right)\sin \left( 6x \right)=-0.35$?

Answer 1

Hint: Now we are given with an equation in sin and cos. First we will evaluate the equation with the subtraction formula of sin which is $\sin \left( A-B \right)=\sin A\cos B-\cos A\sin B$ . Then we will take ${{\sin }^{-1}}$ and hence find the value of x.

Complete step by step solution:
Now we are given with an equation in sin and cos.
Now let us first understand the functions $\sin \theta $ and $\cos \theta $
These functions are actually trigonometric ratios. Now in a right angle triangle,
$\sin \theta =\dfrac{\text{opposite side}}{\text{hypotenuse}}$ and $\cos \theta =\dfrac{\text{adjecent side}}{\text{hypotenuse}}$ .
Now for each angle we get the value for sin and cos.
Now the function sin and cos both gives values only between -1 and 1.
Now we have an identity which say $\sin \left( A-B \right)=\sin A\cos B-\cos A\sin B$
Now consider the given equation $\sin \left( 2x \right)\cos \left( 6x \right)-\cos \left( 2x \right)\sin \left( 6x \right)=-0.35$
Now using $\sin \left( A-B \right)=\sin A\cos B-\cos A\sin B$ we can write $\sin \left( 2x \right)\cos \left( 6x \right)-\cos \left( 2x \right)\sin \left( 6x \right)=\sin \left( 2x-6x \right)$
Hence we get the equation as $\sin \left( 2x-6x \right)=-0.35$
Hence we get, $\sin \left( -4x \right)=-0.35$
Now again we know that $\sin \left( -x \right)=-\sin x$
Hence using this we get, $-\sin \left( 4x \right)=-0.35$
Now multiplying the equation by – 1 we get, $\sin \left( 4x \right)=0.35$
Now we get, $4x={{\sin }^{-1}}\left( 0.35 \right)$ .
Dividing the whole equation by 4 we get,
$x=\dfrac{{{\sin }^{-1}}\left( 0.35 \right)}{4}$ .

Note:
Now note that the formula for $\sin \left( A-B \right)$ is $\sin A\cos B-\cos A\sin B$ hence take care of which value is taken as A or B as this may change the overall sign of the expression. Also for addition we have $\sin \left( A+B \right)=\sin A\cos B+\cos A\sin B$ .