If sinx= 2/3, then find the value of cos 3x.

sinx =  \(\frac{2 }{3}\)

sin²x + cos²x = 1

cos²x = 1 -  \(\frac{4 }{9}\)

= \(\frac{5 }{9}\)

cosx = \(\frac{√5 }{3}\)

We know,

cos3x = 4cos³x - 3cosx

= 4 × ( \(\frac{√5 }{3}\) )³ - 3 × \(\frac{√5 }{3}\)

= \(\frac{20√5 }{27}\) - √5

= - \(\frac{7√5 }{27}\)

= - 0.5797