Find all solutions for the following equations. Write your answers in radians

Question

sin 2x cos x + cos 2x sin x = -1

Answer 1

x = pi/2 + 2npi

x = (1/2 +2n)pi where n = any integer

for n=0

sin(pi)cos(pi/2) + cos(pi)sin(pi/2) = (0)(0) +(-1)(1) = -1

also x = 7pi/6 + 2npi and 11pi/6 + 2npi where n= any integer

for n=0 x=7pi/6+2(0)pi = 7pi/6

sin2xcosx + cos2xsinx

=sin(7pi/3)cos(7pi/6) + cos(7pi/3)sin(7pi/6)

= sin420cos210 + cos420sin210

=sin60(-cos30) + cos60(-sin30)

= sqr3/2(-sqr3/2) -1/4

= -3/4-1/4

= -1

sin2xcosx + cos2xsinx = -1

2sinxcosxcox + (cos^2(x) - sin^2(x))sinx = -1

2sinxcos^2(x)+ (cos^2(x) + sin^2(x))sinx = -1 + 2sin^3(x)

2sin(1-sin^2(x))+ sinx =2sin^3(x) -1

4sin^3(x) -3sinx -1 = 0

for x=90 degree = pi/2

4(1)^3 -3(1) -1 = 4-3-1 = 0

let y= sinx

then

4y^3 -3y -1 = 0

divide that cubic by y-1 to get

4y^2 +4y +1 = 0

(2y+1)^2 = 0

2y+1 = 0

y =-1/2

sinx = -1/2

x = 7pi/6, 11pi/6

x = 7pi/6 +2npi, 11pi/6+2npi where n = any integer