Inequality of trigonometric functions

Question:
(1) The range of $x$ is: $0^o \leqq x \leqq 180^o$. Solve $\cos^4 x > \sin^4 x$.
(2) The range of $x$ is: $0^o \leqq x < 360^o$. Solve $\cos 2x + \sin x < 0$.

Answer:
(1) Arrange the equation.
\begin{eqnarray}
\cos^4 x - \sin^4 x &>& 0  \\
(\cos^2 x + \sin^2 x)(\cos^2 x - \sin^2 x) &>& 0  \\
(1)(\cos x + \sin x)(\cos x - \sin x) &>& 0
\end{eqnarray}
Therefore, we can state
\begin{eqnarray}
& & (\cos x + \sin x)>0 \ \mathrm{and} \ (\cos x - \sin x)>0  \\
\mathrm{or}  & & \nonumber \\
& & (\cos x + \sin x)<0 \ \mathrm{and} \ (\cos x - \sin x)<0
\end{eqnarray}
In case of (4), within the $x$ range, when $\cos x > \sin x$, $0^o \leqq x < 45^o$. When $\cos x + \sin x > 0$, $0^o \leqq x < 135^o$ since the solution of $\cos x + \sin x = 0$ is $x=135^o$. Therefore, one range is
\[
0^o \leqq x < 45^o
\]
In case of (5), when $\cos x < \sin x$, $45^o < x \leqq 180^o$. When $\cos x + \sin x < 0$, $135^o < x \leqq 180^o$. Therefore, the other range is
\[
135^o < x \leqq 180^o
\]
Thus, the answer is $0^o \leqq x < 45^o$ and $135^o < x \leqq 180^o$.

(2) Use the double angle formula, $\cos 2x = 1-2\sin^2 x$ to rearrange the given expression.
\begin{eqnarray}
1-2\sin^2 x +\sin x &<& 0 \\
2\sin^2 x -\sin x-1 &>& 0 \\
(2\sin x + 1)(\sin x -1) &>& 0
\end{eqnarray}
$\sin x -1$ is always negative or zero, so the condition must be only
\[
2\sin x + 1 <0 \ \mathrm{and} \ \sin x -1<0
\]
Then, $2\sin x + 1 <0$ determines the range. Namely, $\sin x < -\frac{1}{2}$. For $\sin x = -\frac{1}{2}$, $x= -30^o$ or $210^o$. Therefore, the range must be
\[
210^o < x < 330^o
\]

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