I am trying to solve $x$ for $\sin^2(x) - \cos(x) - 1 = 0$, for $0°\leqslant x \lt 360°$.
I have the key with the answer $0°$ but have been unable to confirm this using Wolfram Alpha (I assume I have not done the input correctly) and have also not been able to reach it myself. Which steps should I take to reach that answer?
$\endgroup$ 55 Answers
$\begingroup$Use $\displaystyle\sin^2x+\cos^2x=1\iff \sin^2x-1=-\cos^2x$
$$\cos x=0\implies x=(2n+1)90^\circ$$
$$\cos x=-1\implies x=(2m+1)180^\circ$$ where where $m,n$ are integers
I leave for you to find suitable values of $m,n$ so that $x$ lies in the range specified.
$\endgroup$ 4 $\begingroup$Let $t=\cos x$, then using $\sin^2 x + \cos^2 x = 1$, we have $1-t^2 -t -1 = 0 $, or $t^2+t = t(t+1) = 0$.
Hence the solutions are $\arccos 0, \arccos(-1)$ that lie in the desired range.
$\endgroup$ $\begingroup$So we need to solve $\sin^{2}(x) - \cos(x) - 1 = 0$. You should recognize that we have a 1 in this equation, and also a $\sin^{2}(x)$. The trig identity $\sin^{2}(x) + \cos^{2}(x) = 1$ should be jumping out at you. But we would need to manipulate this identity to involve a $\sin^{2}(x)$ and $1$ on one side of the equal sign. To do this, subtract $\sin^{2}(x)$ from both sides, and this gives the identity $\cos^{2}(x) = 1 - \sin^{2}(x)$, or, $-\cos^{2}(x) = \sin^{2}(x) - 1$.
So we want to solve $\sin^{2}(x) - \cos(x) - 1 = 0$, and if we get $\cos(x)$ on one side of the equation, we have to solve $\sin^{2}(x) - 1 = \cos(x)$. But we established above from the $\sin^{2}(x) + \cos^{2}(x) = 1$ identity that $-\cos^{2}(x) = \sin^{2}(x) - 1$, and plugging this into our problem gives that we need to solve:
$$-\cos^{2}(x) = \cos(x)$$
or
$$\cos(x) + \cos^{2}(x) = 0$$
and factoring out a common factor of $\cos(x)$ gives
$$ \cos(x)(1 + \cos(x)) = 0.$$
But $\cos(x)(1 + \cos(x)) = 0$ if either $\cos(x) = 0$ or $1 + \cos(x) = 0$.
So we need to solve two equations. The first is $\cos(x) = 0$. Which angles between $0^{\circ}$ and $360^{\circ}$ give cosine of that angle is 0? Hopefully you said $90^{\circ}$ and $270^{\circ}$.
We also need to solve $1 + \cos(x) = 0$, or $\cos(x) = -1$. Which angles between $0^{\circ}$ and $360^{\circ}$ give cosine of that angle is -1? Hopefully you said $180^{\circ}$.
So your three answers to the equation are $x = 90^{\circ}, 180^{\circ},$ and $270^{\circ}$.
$\endgroup$ $\begingroup$Another approach :\begin{align} \sin^2x-\cos x-\color{red}{1}&=0\\ \sin^2x-\cos x-(\color{red}{\sin^2x+\cos^2x})&=0\\ \cos^2x+\cos x&=0\\ (\cos x+1)\cos x&=0 \end{align} then $\cos x=0\;\color{blue}{\Rightarrow}\;x=(2n+1)\cdot90^\circ$ and $\cos x=-1\;\color{blue}{\Rightarrow}\;x=(2n+1)\cdot180^\circ$, for $n\in\mathbb{Z}$.
$\endgroup$ $\begingroup$considering that $1-sin^2 x =cos^2 x$ you can simplify this to an equation contains only $cos x$
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