Solve the equation algebraically, and then find the values on the interval [0, 2π). (Enter your answers as a comma-separated list. Round to four decimal places.) 6 sin^2(x) + sin(2x) sec(x) − 6 = 0

Answer 1

6sin²x + sin(2x)secx - 6 = 0

6sin²x - 2sinxcosx(1/cosx) - 6 = 0

6sin²x - 2sinx - 6 = 0

3sin²x - sinx - 3 = 0

sinx = [1 ± √37] / 6

sinx = 1.18 or -0.8471

Since sinx can't be bigger than 1, we have sinx = -0.8471

x = Sin^-1(0.8471) + π or 2π - Sin^-1(0.8471) = 4.1521 or 5.2727

Answer 2

6 sin^2(x) + sin(2x) sec(x) − 6 = 0

The best I can get is 58°=.011

I rewrote it: 6*(sinx)^2+s(2x)/cosx-6=0