Question

How do you solve 0=sin^2 x

Answer 1

@SolomonZelman can you help me? :)

Answer 2

\(\normalsize\color{blue}{ 0=\sin^2 x }\)
\(\normalsize\color{blue}{ 0=1-\cos^2 x }\)
\(\normalsize\color{blue}{ \cos^2 x=1 }\)
\(\normalsize\color{blue}{ \cos x= ± 1 }\)

but `│ cos θ│ < 1` so no solution

Answer 3

`│ cos θ│ < 1` because `cos=Adjacent/Hypotenuse` and `Hyp>Adj`

Answer 4

i forgot to mention this but the solutions of the equation have to be in the interval (0, 2pi). would there be a solution now?

Answer 5

You can also know that there is no solution, becuase
\(\normalsize\color{blue}{ \sin^2x=0 }\) gives, \(\normalsize\color{blue}{ \sin x=0 }\) and that gives,
\(\normalsize\color{blue}{ 0= \rm { adjacent ~/~ hypotenuse} }\) but a side can not be equal to zero.

Answer 6

I am just thinking abstractly.

Answer 7

ok thanks!