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Question: How do I solve for the two smallest positive solutions for \(\sin \left( 2x \right)\cos \left( 6x \r...

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

Explanation

Solution

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⁡(A−B)=sin⁡Acos⁡B−cos⁡Asin⁡B\sin \left( A-B \right)=\sin A\cos B-\cos A\sin B . Then we will take sin⁡−1{{\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⁡θ\sin \theta and cos⁡θ\cos \theta
These functions are actually trigonometric ratios. Now in a right angle triangle,
sin⁡θ=opposite sidehypotenuse\sin \theta =\dfrac{\text{opposite side}}{\text{hypotenuse}} and cos⁡θ=adjecent sidehypotenuse\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⁡(A−B)=sin⁡Acos⁡B−cos⁡Asin⁡B\sin \left( A-B \right)=\sin A\cos B-\cos A\sin B
Now consider the given equation sin⁡(2x)cos⁡(6x)−cos⁡(2x)sin⁡(6x)=−0.35\sin \left( 2x \right)\cos \left( 6x \right)-\cos \left( 2x \right)\sin \left( 6x \right)=-0.35
Now using sin⁡(A−B)=sin⁡Acos⁡B−cos⁡Asin⁡B\sin \left( A-B \right)=\sin A\cos B-\cos A\sin B we can write sin⁡(2x)cos⁡(6x)−cos⁡(2x)sin⁡(6x)=sin⁡(2x−6x)\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⁡(2x−6x)=−0.35\sin \left( 2x-6x \right)=-0.35
Hence we get, sin⁡(−4x)=−0.35\sin \left( -4x \right)=-0.35
Now again we know that sin⁡(−x)=−sin⁡x\sin \left( -x \right)=-\sin x
Hence using this we get, −sin⁡(4x)=−0.35-\sin \left( 4x \right)=-0.35
Now multiplying the equation by – 1 we get, sin⁡(4x)=0.35\sin \left( 4x \right)=0.35
Now we get, 4x=sin⁡−1(0.35)4x={{\sin }^{-1}}\left( 0.35 \right) .
Dividing the whole equation by 4 we get,
x=sin⁡−1(0.35)4x=\dfrac{{{\sin }^{-1}}\left( 0.35 \right)}{4} .

Note:
Now note that the formula for sin⁡(A−B)\sin \left( A-B \right) is sin⁡Acos⁡B−cos⁡Asin⁡B\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⁡(A+B)=sin⁡Acos⁡B+cos⁡Asin⁡B\sin \left( A+B \right)=\sin A\cos B+\cos A\sin B .