Question

2Cos(x+y)=2Sin(x-y)=1
find Sin^2x+Cos^2x​

Answer 1

Answer:

2(cosx+2cos

2(cosx+2cos 2

2(cosx+2cos 2 x−1)+2sinxcosx.(1+2cosx)−2sinx=0

2(cosx+2cos 2 x−1)+2sinxcosx.(1+2cosx)−2sinx=0or 2(2cos

2(cosx+2cos 2 x−1)+2sinxcosx.(1+2cosx)−2sinx=0or 2(2cos 2

2(cosx+2cos 2 x−1)+2sinxcosx.(1+2cosx)−2sinx=0or 2(2cos 2 x+cosx−1)+2sinx(2cos

2(cosx+2cos 2 x−1)+2sinxcosx.(1+2cosx)−2sinx=0or 2(2cos 2 x+cosx−1)+2sinx(2cos 2

2(cosx+2cos 2 x−1)+2sinxcosx.(1+2cosx)−2sinx=0or 2(2cos 2 x+cosx−1)+2sinx(2cos 2 x+cosx−1)=0

2(cosx+2cos 2 x−1)+2sinxcosx.(1+2cosx)−2sinx=0or 2(2cos 2 x+cosx−1)+2sinx(2cos 2 x+cosx−1)=02(1+sinx)(cosx+1)(2cosx−1)=0

2(cosx+2cos 2 x−1)+2sinxcosx.(1+2cosx)−2sinx=0or 2(2cos 2 x+cosx−1)+2sinx(2cos 2 x+cosx−1)=02(1+sinx)(cosx+1)(2cosx−1)=0We have to determine values of x s.t. −π≤x≤π

2(cosx+2cos 2 x−1)+2sinxcosx.(1+2cosx)−2sinx=0or 2(2cos 2 x+cosx−1)+2sinx(2cos 2 x+cosx−1)=02(1+sinx)(cosx+1)(2cosx−1)=0We have to determine values of x s.t. −π≤x≤π1+sinx=0 ∴sinx=−1

2(cosx+2cos 2 x−1)+2sinxcosx.(1+2cosx)−2sinx=0or 2(2cos 2 x+cosx−1)+2sinx(2cos 2 x+cosx−1)=02(1+sinx)(cosx+1)(2cosx−1)=0We have to determine values of x s.t. −π≤x≤π1+sinx=0 ∴sinx=−1∴x=2nπ+

2(cosx+2cos 2 x−1)+2sinxcosx.(1+2cosx)−2sinx=0or 2(2cos 2 x+cosx−1)+2sinx(2cos 2 x+cosx−1)=02(1+sinx)(cosx+1)(2cosx−1)=0We have to determine values of x s.t. −π≤x≤π1+sinx=0 ∴sinx=−1∴x=2nπ+ 2

2(cosx+2cos 2 x−1)+2sinxcosx.(1+2cosx)−2sinx=0or 2(2cos 2 x+cosx−1)+2sinx(2cos 2 x+cosx−1)=02(1+sinx)(cosx+1)(2cosx−1)=0We have to determine values of x s.t. −π≤x≤π1+sinx=0 ∴sinx=−1∴x=2nπ+ 23π

2(cosx+2cos 2 x−1)+2sinxcosx.(1+2cosx)−2sinx=0or 2(2cos 2 x+cosx−1)+2sinx(2cos 2 x+cosx−1)=02(1+sinx)(cosx+1)(2cosx−1)=0We have to determine values of x s.t. −π≤x≤π1+sinx=0 ∴sinx=−1∴x=2nπ+ 23π

2(cosx+2cos 2 x−1)+2sinxcosx.(1+2cosx)−2sinx=0or 2(2cos 2 x+cosx−1)+2sinx(2cos 2 x+cosx−1)=02(1+sinx)(cosx+1)(2cosx−1)=0We have to determine values of x s.t. −π≤x≤π1+sinx=0 ∴sinx=−1∴x=2nπ+ 23π ∴x=−

2(cosx+2cos 2 x−1)+2sinxcosx.(1+2cosx)−2sinx=0or 2(2cos 2 x+cosx−1)+2sinx(2cos 2 x+cosx−1)=02(1+sinx)(cosx+1)(2cosx−1)=0We have to determine values of x s.t. −π≤x≤π1+sinx=0 ∴sinx=−1∴x=2nπ+ 23π ∴x=− 2

2(cosx+2cos 2 x−1)+2sinxcosx.(1+2cosx)−2sinx=0or 2(2cos 2 x+cosx−1)+2sinx(2cos 2 x+cosx−1)=02(1+sinx)(cosx+1)(2cosx−1)=0We have to determine values of x s.t. −π≤x≤π1+sinx=0 ∴sinx=−1∴x=2nπ+ 23π ∴x=− 2π

2(cosx+2cos 2 x−1)+2sinxcosx.(1+2cosx)−2sinx=0or 2(2cos 2 x+cosx−1)+2sinx(2cos 2 x+cosx−1)=02(1+sinx)(cosx+1)(2cosx−1)=0We have to determine values of x s.t. −π≤x≤π1+sinx=0 ∴sinx=−1∴x=2nπ+ 23π ∴x=− 2π

2(cosx+2cos 2 x−1)+2sinxcosx.(1+2cosx)−2sinx=0or 2(2cos 2 x+cosx−1)+2sinx(2cos 2 x+cosx−1)=02(1+sinx)(cosx+1)(2cosx−1)=0We have to determine values of x s.t. −π≤x≤π1+sinx=0 ∴sinx=−1∴x=2nπ+ 23π ∴x=− 2π ,

2(cosx+2cos 2 x−1)+2sinxcosx.(1+2cosx)−2sinx=0or 2(2cos 2 x+cosx−1)+2sinx(2cos 2 x+cosx−1)=02(1+sinx)(cosx+1)(2cosx−1)=0We have to determine values of x s.t. −π≤x≤π1+sinx=0 ∴sinx=−1∴x=2nπ+ 23π ∴x=− 2π ,for n=−1 ∵−π≤x<π

2(cosx+2cos 2 x−1)+2sinxcosx.(1+2cosx)−2sinx=0or 2(2cos 2 x+cosx−1)+2sinx(2cos 2 x+cosx−1)=02(1+sinx)(cosx+1)(2cosx−1)=0We have to determine values of x s.t. −π≤x≤π1+sinx=0 ∴sinx=−1∴x=2nπ+ 23π ∴x=− 2π ,for n=−1 ∵−π≤x<πcosx=−1=cosπ ∴cosx=1/2=cos(π/3)

2(cosx+2cos 2 x−1)+2sinxcosx.(1+2cosx)−2sinx=0or 2(2cos 2 x+cosx−1)+2sinx(2cos 2 x+cosx−1)=02(1+sinx)(cosx+1)(2cosx−1)=0We have to determine values of x s.t. −π≤x≤π1+sinx=0 ∴sinx=−1∴x=2nπ+ 23π ∴x=− 2π ,for n=−1 ∵−π≤x<πcosx=−1=cosπ ∴cosx=1/2=cos(π/3)∴x=2nπ±π/3

2(cosx+2cos 2 x−1)+2sinxcosx.(1+2cosx)−2sinx=0or 2(2cos 2 x+cosx−1)+2sinx(2cos 2 x+cosx−1)=02(1+sinx)(cosx+1)(2cosx−1)=0We have to determine values of x s.t. −π≤x≤π1+sinx=0 ∴sinx=−1∴x=2nπ+ 23π ∴x=− 2π ,for n=−1 ∵−π≤x<πcosx=−1=cosπ ∴cosx=1/2=cos(π/3)∴x=2nπ±π/3∴x=π/3, −π/3

2(cosx+2cos 2 x−1)+2sinxcosx.(1+2cosx)−2sinx=0or 2(2cos 2 x+cosx−1)+2sinx(2cos 2 x+cosx−1)=02(1+sinx)(cosx+1)(2cosx−1)=0We have to determine values of x s.t. −π≤x≤π1+sinx=0 ∴sinx=−1∴x=2nπ+ 23π ∴x=− 2π ,for n=−1 ∵−π≤x<πcosx=−1=cosπ ∴cosx=1/2=cos(π/3)∴x=2nπ±π/3∴x=π/3, −π/3Hence the values of x s.t. −π≤x≤π are

2(cosx+2cos 2 x−1)+2sinxcosx.(1+2cosx)−2sinx=0or 2(2cos 2 x+cosx−1)+2sinx(2cos 2 x+cosx−1)=02(1+sinx)(cosx+1)(2cosx−1)=0We have to determine values of x s.t. −π≤x≤π1+sinx=0 ∴sinx=−1∴x=2nπ+ 23π ∴x=− 2π ,for n=−1 ∵−π≤x<πcosx=−1=cosπ ∴cosx=1/2=cos(π/3)∴x=2nπ±π/3∴x=π/3, −π/3Hence the values of x s.t. −π≤x≤π are−π,−π/2,−π/3,π/3,π.

Step-by-step explanation:

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