What is the value of x in the following equation fo [0,2Pi]? Sin2x = - cos2x

Question

384 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-04-04T02:40:05

#x={(3pi)/8,(7pi)/8,(11pi)/8,(15pi)/8}#

As #sin2x=-cos2x#

#(sin2x)/(cos2x)=-1#

i.e. #tan2x=tan(-pi/4)#

and #2x=npi-pi/4#

or #x=(npi)/2-pi/8#

and in the interval #[0,2pi]#, we can have

#x={(3pi)/8,(7pi)/8,(11pi)/8,(15pi)/8}#