6. Solve the following equations on the interval x
a) 6 cos? * + cosx - 1 = 0
b) 8 cos? x + 14 cosx
- 3
c) sec? x + 2 secx - 8 = 0
a) To solve the equation 6cos^2(x) + cos(x) - 1 = 0, let's substitute u = cos(x) to simplify the equation:
6u^2 + u - 1 = 0
Now, we can factor the quadratic equation:
(3u + 1)(2u - 1) = 0
Setting each factor to zero gives us the solutions:
3u + 1 = 0 or 2u - 1 = 0
u = -1/3 or u = 1/2
Since u = cos(x), the solutions for x are:
cos(x) = -1/3 or cos(x) = 1/2
The solutions for x on the interval [0, 2π] are:
x = arccos(-1/3) or x = arccos(1/2)
b) To solve the equation 8cos^2(x) + 14cos(x) - 3 = 0, let's substitute u = cos(x) to simplify the equation:
8u^2 + 14u - 3 = 0
Now, we can factor the quadratic equation:
(4u - 1)(2u + 3) = 0
Setting each factor to zero gives us the solutions:
4u - 1 = 0 or 2u + 3 = 0
u = 1/4 or u = -3/2
Since u = cos(x), the solutions for x are:
cos(x) = 1/4 or cos(x) = -3/2
The solutions for x on the interval [0, 2π] are:
x = arccos(1/4) or x = arccos(-3/2)
c) To solve the equation sec^2(x) + 2sec(x) - 8 = 0, let's substitute u = sec(x) to simplify the equation:
u^2 + 2u - 8 = 0
Now, we can factor the quadratic equation:
(u + 4)(u - 2) = 0
Setting each factor to zero gives us the solutions:
u + 4 = 0 or u - 2 = 0
u = -4 or u = 2
Since u = sec(x), the solutions for x are:
sec(x) = -4 or sec(x) = 2
The solutions for x on the interval [0, 2π] are:
x = arccos(-1/4) or x = arccos(1/2)