Question

How would you solve this equation?

Answer 1

`beta = 120^@, 240^@ " or " (2pi)/3, (4pi)/3`

Explanation:

`2cos^2 beta - 7 cos beta - 4 = 0`

`2 cos^2 beta + cos beta - 8 cos beta - 4 = 0`

`cos beta * (2 cos beta + 1) - 4 * (2cos beta + 1) = 0`

`(cos beta -4) * (2 cos beta + 1) = 0`

`cos beta = 4, -1/2`

Since cos can have a value between -1 & +1 only, 4 is not a valid solution.

`cos beta = -1/2`

Cosine is negative in II and III Quadrant.

`:. beta = 120^@, 240^@`

Answer 2

`beta = 120^@, 240^@`.

Explanation:

This will be a piece of cake if you just assume a variable equal to `cos beta`.

So, Let's Assume `cos beta = x`.

Now the Equation is,

`2x^2 - 7x - 4 = 0`

`rArr 2x^2 -(8 - 1)x - 4 = 0` [Well, `7 = 8 -1`, right?]

`rArr 2x^2 - 8x + x - 4 = 0` [Distributive Property]

`rArr 2x(x - 4) + 1 (x - 4) = 0` [Grouping Like Terms]

`rArr (x - 4)(2x + 1) = 0` [Grouping Again.]

So, Either `x - 4 = 0 rArr x = 4`

Or, `2x - 1 = 0 rArr 2x = -1 rArr x = -1/2`

So, `cos beta = 4, -1/2`.

Now,

Cosine (`cos`) is a periodic function whose value ranges from `-1` to `1`. [`-1 <= cos x <= 1`]

So, `cos beta = -1/2`

Now, `beta = 120^@, 240^@`. [As `beta = [0^@, 360^@)`]

Hope this helps.