6cos^2x + sinx = 4

Question

ended up with sinx (6sinx -1)=2

Answer 1

6cos^2(x) +sin(x) = 4

replace cos^2(x) with 1-sin^2(x)

6(1-sin^2(x)) +sin(x) -4 = 0

6 -6sin^2(x) +sin(x) -4 = 0

-6sin^2(x) + sin(x) +2 = 0 multiply by -1

6sin^2(x) - sin(x) -2 = 0 factor

(3sin(x) - 2)(2sin(x) +1) = 0 set each factor =0 and solve for sin(x)

sin(x) = 2/3 or sin(x) =-1/2 use a calculator with an inverse sine function

x = about 41.81 degrees or 180-41.81 = about 138.19 degrees

or x = 330 degrees or 210 degrees, or in radians x = 11pi/6 or 7pi/6

x = 41.81, 138.19, 210 or 330 degrees

or x = 41.81pi/180, 138.19pi/180, 11pi/6, 7pi/6 radians

x = about .23pi, .77pi, 11pi/6, 7pi/6

it's about pi/4, 3pi/4 and exactly 11pi/6, 7pi/6

if you're restricting the answer to 0 to 2pi or 0 to 360 degrees

Answer 2

6( 1 - sin^2 x ) + sin x - 4 = 0 (now move everything to other side)

=> 6sin^2 x - sin x - 2 = 0. (Now factor)

=> ( 2 sin x - 1) ( 3 sin x + 2) = 0

So, 2 sin x - 1 = 0. Or 3 sin x + 2 = 0

=> sin x = 1/2. Or sin x = - 2/ 3

Thus the solution is arcsin (1/2), arcsin (- 2/3) and you can find the corresponding angles based on the interval that you are given.

Patrick

Answer 3

6 - 6sin²x + sinx = 4; 6sin²x - sinx - 2 = 0; (2sinx + 1)(3sinx - 2) = 0;

sinx = -1/2 or sinx = 2/3

x = 7π/6 + 2πn, x = 11π/6 + 2πn; x = sin^-1(2/3) + 2πn, x = π - sin^-1(2/3) + 2πn; n⊂Z