Find the exact solutions to the given equations in the inter

Find the exact solutions to the given equations in the interval [0,2pi):      A) sin^2= cosx+1      B) cos2x+ 3cosx-1= 0      C) csc^2+ cscx= 2      D) tan x/2= sinx     E) sinx+ cosx= 1      F) sinxcosx= -1/2

Solution

interval [0,2pi)

A) sin^2x = cosx+1   

Use identity : sin^2x +cos^2x =1

1-cos^2x = cosx +1

cos^2x +cosx =0

cosx( cosx +1) =0 ---->

cosx =0

x = pi/2 , 3pi/2

cosx = -1

x = pi

Solution : pi/2 , pi , 3pi/2

B) cos2x+ 3cosx-1= 0   

use identity : cos2x = 2cos^2x -1

So,2cos^2x -1 +3cosx -1 =0

2cos^2x +3cosx -2 =0

2cos^2x +4cosx -cosx -2 =0

2cosx( cosx +2) -1(cosx +2) =0

(2cosx -1)(cosx+2) =0

neglect : cosx +2 =0

so, 2cosx -1 =0

cosx = 1/2

x = pi/3 , 2pi -pi/3 = pi/3 , 5pi/3

   C) csc^2+ cscx= 2     

csc^2x + cscx -2 =0

csc^2x + 2cscx -cscx -2 =0

cscx( cscx +2) -1(cscx +2)=0

(cscx +2)(cscx -1) =0 ;

cscx = -2 ----> sinx = -1/2

x = pi +pi/6 , 2pi -pi/6

x = 7pi/6 , 11pi/6

cscx -1 =0

sinx = 1

x = pi/2

Solution : x = pi/2 , 7pi/6 , 11pi/6

D) tan x/2= sinx   

sinx/2 /cosx/2 = 2sinx/2*cosx/2

Now sinx/2 /cosx/2 - 2sinx/2*cosx/2 =0

sin(x/2) ( 1-cos^2(x/2) =0

sin(x/2) =0--->

x/2 = pi ---> x = 2pi ( outside the given interval)

1- cos^2(x/2) =0

cos(x/2) = 1 ----> x/2 = 0 ,

x =0

cos(x/2) = -1 ----> x/2 =pi

x = 2pi ( outside the given interval)

Solution : x = 0