Question

3a)Use the substitutions 𝑥 = 2𝑡𝑎𝑛𝜃, 0 < θ < π2 to express √4 + 𝑥2 as a trig function of θ? 3b) Solve for the general solutions of the trigonometric equation 𝑠𝑖𝑛𝑥 + √2 = −𝑠𝑖𝑛𝑥

Answer 1

For a:
We can substitute in the given value for `x`:
`sqrt(4 + x^2) = sqrt(4 + (2tantheta)^2) = sqrt(4(1 + tan^2theta))`
`= 2 * sqrt(1 + tan^2theta) = 2 sqrt(sec^2(theta)) = 2 |sec(theta)|`
Since `theta in (0, pi/2)`, `sec(x) > 0` so we can just write that it simplifies to
`sqrt(4+x^2) = 2 sec(theta)`

For b:
`sin(x) + sqrt(2) = - sin(x)`
`2sin(x) = -sqrt2`
`sin(x) = - sqrt(2)/2`

Using the unit circle we see that this happens at
`x = (5pi)/4, (7pi)/4`

and then every revolution past, i.e. `+2pi n` for integer `n`:

`x in { (5pi)/4 + 2pi n, (7pi)/4 + 2pi n } forall n in mathbb(Z)`