Question

Find the principal solution and general solution of the following:
sin θ = ${\displaystyle -\frac{1}{\sqrt{2}}}$

Answer 1

sin θ = ${\displaystyle -\frac{1}{\sqrt{2}}}$

We know that principal of sin θ lies in ${\displaystyle [-\frac{\pi }{2},\frac{\pi }{2}]}$

sin θ = ${\displaystyle -\frac{1}{\sqrt{2<0}}}$

∴ The principal value of sin θ lies in the IV quadrant.

sin θ = ${\displaystyle -\frac{1}{\sqrt{2}}}$

= ${\displaystyle -\sin \left(\frac{\pi }{4}\right)}$

sin θ = ${\displaystyle \sin (-\frac{\pi }{4})}$

Hence θ = ${\displaystyle -\frac{\pi }{4}}$ is the principal solution.

The general solution is

θ = nπ + (– 1)ⁿ . ${\displaystyle \left(\frac{\pi }{4}\right)}$, n ∈ Z

θ = ${\displaystyle \text{n}\pi +{(-1)}^{\text{n}+1}\cdot \frac{\pi }{4}}$, n ∈ Z