Solve the equation 2secxsinx+2=4sinx+secx2secxsinx+2=4sinx+secx in the interval [0^@,360^@][0,360]?

1 Answer

x=129.564^@x=129.564 and x=321.564^@x=321.564

Explanation:

2sec xsin x + 2 = 4sin x + sec x2secxsinx+2=4sinx+secx

(2sin x)/(cos x) + 2 = (4sinxcos x + 1)/(cos x)2sinxcosx+2=4sinxcosx+1cosx

2(sin x + cos x) = 4sinxcos x + 12(sinx+cosx)=4sinxcosx+1 (1)
- Multiplying both side by cos x (condition cos x diff. to zero)

Call (sin x + cos x ) = u(sinx+cosx)=u

u^2 = (sin x + cos x)^2 = sin^2 x + cos^2 x + 2sinxcos x u2=(sinx+cosx)2=sin2x+cos2x+2sinxcosx

= 1 + 2sin xcos x=1+2sinxcosx.

2sin xcos x = u^2 - 12sinxcosx=u21

4sin xcos x = 2u^2 - 24sinxcosx=2u22

Substitute these values into (1):

2u = 2u^2 - 2 + 12u=2u22+1

2u^2 - 2u - 1 = 02u22u1=0

Solve this quadratic equation for u = (sin x +cos x)u=(sinx+cosx)
D = d^2 = b^2 - 4ac = 4 + 8 = 12D=d2=b24ac=4+8=12 --> d = +- 2sqrt3d=±23

There are 2 real roots:
sin x + cos x = u = 2/2 +- (2sqrt3)/4 = 2 +- sqrt3/2sinx+cosx=u=22±234=2±32
u_1 = 1 + sqrt3/2u1=1+32 (Rejected as >sqrt2>2)
u_2 = 1 - sqrt3/2 = 0.134u2=132=0.134

sin x + cos x = sqrt2cos (x - pi/4) = 0.134sinx+cosx=2cos(xπ4)=0.134
cos (x - pi/4) = 0.134/sqrt2 = 0.0947cos(xπ4)=0.1342=0.0947

x - pi/4 = +- 84.564^@xπ4=±84.564
x = 84.564 + 45 = 129.564^@x=84.564+45=129.564 and
x = 360- 84.564 + 45 = 321.564^@x=36084.564+45=321.564