Question

Precalculus PLEASE HELP ME right answers only check all please

Answer 1

Given :

`cos(x)cos((\pi )/(7)) + sin(x)sin((\pi )/(7))=-(√(2))/(2)`

We know the identity

`cos(x)cos(y) + sin(x)sin(y)=cos(x-y)`

We use this property to simplify the left hand side

So
`cos(x)cos((\pi )/(7)) + sin(x)sin((\pi )/(7))=cos(x - (\pi )/(7))`

`cos(x - (\pi )/(7)) =-(√(2))/(2))`

we know ,

when
`cos(x) =-(√(2))/(2))` then

`x = (3\pi )/(4) , (5\pi )/(4)`

For
`cos(x - (\pi )/(7)) =-(√(2))/(2))`

`x - (\pi )/(7)= (3\pi )/(4) and x - (\pi )/(7)= (5\pi )/(4)`

Add
`(\pi )/(7)` on both sides

`x= (3\pi )/(4) + (\pi )/(7) and x= (5\pi )/(4) +(\pi )/(7)`

Finally we add 2npi for general solution

So options C and D are correct