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The equation is:
`3cos^2 x + cos x - 1 = 0`
Notice that this equation resembles a quadratic equation. Hence, it will be easier for us to solve this if we transform it first to a quadratic equation, which we can solve by using the quadratic formula.
We first make the assignment `u = cos x` .
Hence, the equation becomes:
`3u^2 + u - 1 = 0` to which we apply the quadratic formula:
`u = (-1 pm sqrt(1^2 - 4*3*-1))/(2*3)`
This gives us:
`u = (-1 pm sqrt(13))/6`
or `u approx 0.434 ; u approx -0.768`
Now, we transform this back to cosine. We know that `u = cos x` .
This means that:
`cos x = 0.434` or `cos x = -0.768`
Applying `arccos` on both sides:
`x = cos^-1(0.434) = 1.12`
`x = cos^-1(-0.768) = 2.45`
Hence, the correct answer is b 1.12 and 2.45.
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