Question

How do you solve `cotx=tan(2x-3pi)` in the interval [0,360]?

Answer 1

`x={30^o,150^o,210^o,330^o}`

Explanation:

`cotx=tan(2x-3pi)=tan-(3pi-2x)=-tan(3pi-2x)=-(-tan2x)=tan2x`

Hence `1/tanx=(2tanx)/(1-tan^2x)` or

`2tan^2x=1-tan^2x` or

`3tan^2x=1` or

`tan^2x=1/3` and

`tanx=+-1/sqrt3=+-tan(pi/6)=tan(+-pi/6)`

Hence `x=npi+-pi/6`

and in interval `[0^o,360^o]`

`x={30^o,150^o,210^o,330^o}`

Answer 2

`"Solution set "{30^@,90^@,150^@,210^@,270^@,330^@}`

Explanation:

`cotx=tan(2x-3pi)`

`=>tan(2x-3pi)=cotx`

`=>tan(-3pi+2x)=tan(pi/2-x)`

`=>tan(2x)=tan(pi/2-x)`

`=>2x=npi+pi/2-x," where n" in ZZ`

`=>3x=npi+pi/2," where n" in ZZ`

`=>x=1/3xxnpi+pi/6," where n" in ZZ`

`=>x=1/6(2n+1)pi" where n" in ZZ`

To evaluate x in the interval `[0.360]`

For n = 0

`=>x=1/6(2xx0+1)pi=180/6=30^@`

For n =1

`=>x=1/6(2xx1+1)pi=180/2=90^@`

For n =2

`=>x=1/6(2xx2+1)pi=(5xx180)/6=150^@`

For n =3

`=>x=1/6(2xx3+1)pi=(7xx180)/6=210^@`

For n =4

`=>x=1/6(2xx4+1)pi=(9xx180)/6=270^@`

For n =5

`=>x=1/6(2xx5+1)pi=(11xx180)/6=330^@`