Answer:
The solution is [tex]x =0.[/tex]
Step-by-step explanation:
The equation
[tex]\bold{(1).}\: sin(\dfrac{3\pi }{2} +x)+sin(\dfrac{3\pi }{2} +x) =-2[/tex]
can be rewritten as
[tex]2* sin(\dfrac{3\pi }{2} +2) =-2[/tex]
and can be further simplified to
[tex]sin(\dfrac{3\pi }{2}+x ) =-1[/tex].
Now, taking the inverse sine of both sides we get:
[tex]sin^{-1}(sin(\dfrac{3\pi }{2} +x) =sin^{-1}(-1)[/tex]
[tex]\bold{(2).}\: \dfrac{3\pi }{2} +2 =sin^{-1}(-1)[/tex]
The value of the right side on the interval [tex][0,2\pi)[/tex] is
[tex]sin^{-1}(-1) = \dfrac{3\pi}{2} \: \:( \text{or }270^o)[/tex] ,
which makes the equation (2)
[tex]\dfrac{3\pi }{2}+x =\dfrac{3\pi}{2}[/tex]
solving for [tex]x[/tex] gives
[tex]\boxed{x = 0}[/tex]
which is our solution.
