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r-ruslan [8.4K]
3 years ago
14

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

Mathematics
2 answers:
Arturiano [62]3 years ago
6 0

The solution would be like this for this specific problem:

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

<span>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.</span>

NeX [460]3 years ago
6 0

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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