Solve algebraically: `2sinxcosx=cosx` Solve algebraically: `sinx/2=sinx/3` solve algeibraically:`cscx/5+cscx/3=16/15`

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(1) 2sinxcosx=cosx 

** Do not divide by cosx; you may lose a root **


cosx(2sinx-1)=0 By the zero product property:

cosx=0 ==> `x=pi/2+npi,n in ZZ` (n an integer)

`2sinx-1=0 ==> sinx=1/2 ==> x=pi/6+2npi,x=(5pi)/6+2npi` So the solutions are `x=pi/2+npi,x=pi/6+2npi,x=(5pi)/6+2npi`

(2) `(sinx)/2=(sinx)/3`



sinx=0 ==> `x=npi`

(3) `(cscx)/5+(cscx)/3=16/15`



`cscx=2==> sinx=1/2`

`=> x=pi/6+2npi,x=(5pi)/6+2npi`

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