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which of the following are solutions to the equation sinx cos(pi/7)-sin(pi/7)cosx=sqrt(2)/2

check all that apply

A7pi/4+pi/4+2npi
B5pi/4+pi/7+2npi
Cpi/4+pi/7+2npi
D3pi/4+pi/7+2npi

Verified Answer :

The solution would
be like this for this specific problem:

sin ( x - ( pi / 7 ) ) = - sqrt ( 2 ) / 2
x - ( pi / 7 ) = - pi / 4 + 2npi or x - ( pi / 7 ) = (5pi / 4 ) + 2npi
x = ( pi / 7 ) - ( pi / 4 ) + 2npi or x = ( 5pi / 4 ) + ( pi / 7 ) + 2npi
x = ( - 3pi / 28 ) + 2n*pi 

I am hoping that this answer has satisfied your query and it will be able to help you in your endeavor, and if you would like, feel free to ask another question.

Answer:

The correct answer is:

Option: D

D.  3pi/4+pi/7+2npi

Step-by-step explanation:

We know that:

\(\sin A\cos B-\cos A\sin B=\sin(A-B)\)

Here we are given:

\(\sin x\cos (\dfrac{\pi}{7})-\sin (\dfrac{\pi}{7})\cos x=\dfrac{\sqrt{2}}{2}=\dfrac{1}{\sqrt{2}}\)

so, this quantity as by the above formula will be equal to :

\(\sin (x-\dfrac{\pi}{7})=\dfrac{1}{\sqrt{2}}\)

Now, we will check which options or which value of x will hold the following equation true.

Option: D is the correct answer.

i.e. when we put:

\(x=\dfrac{3\pi}{4}+\dfrac{\pi}{7}+2n\pi\)

we get:

\(\sin (\dfrac{3\pi}{4}+\dfrac{\pi}{7}+2n\pi-\dfrac{\pi}{7})=\dfrac{1}{\sqrt{2}}\\\\\sin (2n\pi+\dfrac{3\pi}{4})=\dfrac{1}{\sqrt{2}}\\\\\sin (\dfrac{3\pi}{4})=\dfrac{1}{\sqrt{2}}\\\\\sin (\pi-\dfrac{\pi}{4})=\dfrac{1}{\sqrt{2}}\\\\\sin (\dfrac{\pi}{4})=\dfrac{1}{\sqrt{2}}\\\\\dfrac{1}{\sqrt{2}}=\dfrac{1}{\sqrt{2}}\)

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