Question

2sin²(x)-5cos(x)+1=0

Answer 1

Remember the basic trig identity


`\sin^2 x + \cos^2 x = 1`

We can move the cosine term to the right side to get


`\sin^2 x = 1 - \cos^2 x`

We can then replace
`\sin^2 x` in the original equation and basically factor the resulting equation like a quadratic:


`2(1 - \cos^2 x) - 5 \cos x + 1 = 0`

`(2 - 2 \cos^2 x) - 5 \cos x + 1 = 0`

`{-2} \cos^2 x - 5 \cos x + 3 = 0`

`2 \cos^2 x + 5 \cos x - 3 = 0`

`(2 \cos x - 1)(\cos x + 3) = 0`

Then, we can see that
`\cos x` either equals
`(1)/(2)` or
`-3`. But since the range of cosine is only from
`[-1, 1]`, cosine can't equal
`-3` at all! So, you just have to solve for x when
`\cos x = (1)/(2)`, which is when


`x = \{ (\pi)/(3), (5\pi)/(3) \} + 2 \pi k | k \in \mathbb{Z}`
(The last part of the solution says that k can be any integer.)

If you only want solutions in the range
`[0, 2\pi)`, then your answer is just


`x = (\pi)/(3), (5\pi)/(3)`