Question

How do you solve `sin 2(x)- 2sin(x)- 1 = 0`?

Answer 1

`{pi(1 + 2n) + arcsin(0.2542), 2pi(1 + n) - arcsin(0.2542), pi(1 + 2n) + arcsin(0.8968) and 2pi(1 + n) - arcsin(0.8968)}`

Explanation:

Use the identity `sin2theta = 2sinthetacostheta`.

`2sinxcosx - 2sinx - 1 = 0`

The challenge here is that we are caught with two different trigonometric functions an equation that is not factorable. In other words, we have to get rid of either sine or cosine in the equation. We can do this as follows.

`2sinxcosx = 1 + 2sinx`

`(2sinxcosx)^2 = (1 + 2sinx)^2`

`4sin^2xcos^2x = 1 + 4sinx + 4sin^2x`

`4sin^2x(1- sin^2x) = 1 + 4sinx + 4sin^2x`

`4sin^2x - 4sin^4x = 1 + 4sinx + 4sin^2x`

`0 = 4sin^4x + 4sinx + 1`

We let `t= sinx`:

`0= 4t^4 + 4t + 1`

We solve using a graphing calculator to get `t = -0.2542` and `t = -0.8968`.

So, to put this into context, `sinx = -0.2542` and `sinx = -0.8968`.

Therefore, using our knowledge of sine's positivity and negativity in the various quadrants, we can deduce that `x = pi + arcsin(0.2542) + 2pin, 2pi - arcsin(0.2542) + 2pin, pi + arcsin(0.8968) + 2pin and 2pi - arcsin(0.8968) + 2pin`

These can be simplified as:

`{pi(1 + 2n) + arcsin(0.2542), 2pi(1 + n) - arcsin(0.2542), pi(1 + 2n) + arcsin(0.8968) and 2pi(1 + n) - arcsin(0.8968)}`

Hopefully this helps!