Question

If `cos3x+sin(2x-(7pi)/6)=-2` , then `x` is equal to `(k in Z)` `pi/3(6k+1)` (b) `pi/3(6k-1)` `pi/3(2k+1)` (d) none of these

Answer 1

`cos3xx+sin(2x-7/5pi)=-2`
`1+cos3x+1+sin(3x-7/6pi)=0`
`2cos^2 3/2x+1+sin(3x-2/3pi-pi/2)=0`
`2cos^2(3/2x)+1-sin(pi/2-(-3x+2/3pi))=0`
`2cos^2(3/2x)+1-cos(2/3pi-2x)=0`
`2cos^2(3/2x)+2sin^2(pi/3-x)=0`
`cos3/2x=0,sin(pi/3-x)=0`
`3/2x=pi/2,3/2pi,5/2pi,7/2pi...`
`x=pi/3,pi,5/3pi,7/3pi,....`
`x=(6k+1)pi/3`
Option A is correct.