Question

Question #d3a7a

Answer 1

Solution: `x=31.7^0(1dp) and x=121.7^0(1dp)`

Explanation:

#sin2x=2cos2x or (sin2x)/(cos2x)=2 or tan2x=2 or 2x= tan^-1 2 =63.4349^0:. x=31.7^0(1dp) # and since #tan(pi+2x)=tan2x=2 :. 2x=180+63.4349=243.439^0:.x=121.7(1dp)#

Solution: `x=31.7^0(1dp) and x=121.7^0(1dp)`[Ans]

Answer 2

`sin2x=2cos2x`

`=>tan2x=2=tan^-1(2)=alpha`

So `2x=npi+alpha," where " n in ZZ`

`x=(npi)/2+alpha/2=(npi)/2+31.7^@`

by calculator, `alpha =tan^-1(2)= 63.4^@`