Question

Question #d2a4f

Answer 1

The solutions are:

`x = 2kpi +- arccos((sqrt3-1)/2) color(white)"XXXXX" k in ZZ`

Or in decimal form:

`x = 2kpi +- 1.196 color(white)"XXXXXX" k in ZZ`

Explanation:

Substitute `y` for `cosx`:

`2cos^2x + 2cosx - 1 = 0`

`2y^2 + 2y - 1 = 0`

Now use the quadratic formula. `a=2, b=2, c=-1`:

`y = (-2+-sqrt(2^2 - 4(2)(-1)))/(2(2))`

`y = (-2+-sqrt(12))/4`

`y = (-2+-2sqrt3)/4`

`y = (-1+-sqrt3)/2`

So we have:

`cosx = (-1-sqrt3)/2 " " or " " cosx = (-1+sqrt3)/2`

But, since `(-1-sqrt3)/2` is less than 1, our only solutions can be when:

`cosx = (sqrt3-1)/2`

Since cos(x) is the same as cos(-x), we can say that:

`x = +-arccos((sqrt3-1)/2)`

And we can't forget that rotating angles by `2kpi` for any integer `k` keeps all trig functions the same:

`x = 2kpi +- arccos((sqrt3-1)/2)`

Finally, plugging in `arccos((sqrt3-1)/2)` to a calculator gives:

`x = 2kpi +- 1.196`

Final Answer