#sin(60+x)-sin(60-x)=1# the interval #0<x<2pi# how do you solve?

1 Answer
Feb 17, 2018

#x=pi/2#

Explanation:

We want to solve the equation

#sin(60^@+x)-sin(60^@-x)=1#

We will use the identity

  • #sin(a+b)-sin(a-b)=2sin(b)cos(a)#

(Can be derived using angle-sum and difference identities)

/

Apply the identity (#a=60# and #b=x#)

#2sin(x)cos(60^@)=1#

Solve for sine

#sin(x)=1/(2cos(60^@))#

#=1/(2*1/2)#

#=1#

By the inverse sine

#x=pi/2#