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How would I solve this?

$\sin^2(\theta) + \cos(\theta) =1$ and $0 < \theta < \pi$

$\cos^2(\theta)= 1- \sin^2(\theta)$.

Now what? In the first or second quadrant the cos can be positive or negative oh wait but cos can't be negative because it's not being squared right?

Idk help please?

2 Answers 2

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We have

$$\sin^2(\theta)+\cos(\theta)-1=0$$

which is equivalent to

$$\cos(\theta)(1-\cos(\theta))=0$$

So, solutions are either $\cos(\theta)=0$ or $\cos(\theta)=1$. With $\theta\in (0,\pi)$, the solutions is $\theta =\pi/2$.

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Suppose $\sin^2 \theta +\cos \theta =1$ for $0<\theta<\pi$ (I can not physically do degrees)

Thus $$ \cos\theta=1-\sin^2\theta $$ we know from our identity $\sin^2x+\cos^2 x=1$ so we get

$$ \cos \theta=\cos^2 \theta $$ so when $\theta=\frac{\pi}{2}$ then we get $0=0$ else we divide by $\cos \theta$. $$ 1=\cos \theta $$ so $$ \theta=Arccos(1) $$ So then calculate what values of $\theta$ are $1$.

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    You missed $\theta=\pi/2$ – 2017-02-07
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    I did i shall fix it... – 2017-02-07
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    And $\arccos(1)$ is inadmissible since $\theta \in (0,\pi)$. – 2017-02-07