Question:

(1) The range of \(x\) is: \(0^o \leq x \leq 180^o\). Solve \(\cos^4 x > \sin^4 x\).

(2) The range of \(x\) is: \(0^o \leq x < 360^o\). Solve \(\cos 2x + \sin x < 0\).

Answer:

(1) Arrange the equation.

\[  \cos^4 x – \sin^4 x > 0  \]

\[  (\cos^2 x + \sin^2 x)(\cos^2 x – \sin^2 x) > 0  \]

\[  (1)(\cos x + \sin x)(\cos x – \sin x) > 0  \]

Therefore, we can state

\[   (\cos x + \sin x)>0 \ \mathrm{and} \ (\cos x – \sin x)>0  \]

\[  \mathrm{or}    \]

\[   (\cos x + \sin x)<0 \ \mathrm{and} \ (\cos x – \sin x)<0  \]

In case of (4), within the \(x\) range, when \(\cos x > \sin x\), \(0^o \leq x < 45^o\). When \(\cos x + \sin x > 0\), \(0^o \leq x < 135^o\) since the solution of \(\cos x + \sin x = 0\) is \(x=135^o\). Therefore, one range is

\[  0^o \leq x < 45^o  \]

In case of (5), when \(\cos x < \sin x\), \(45^o < x \leq 180^o\). When \(\cos x + \sin x < 0\), \(135^o < x \leq 180^o\). Therefore, the other range is

\[  135^o < x \leq 180^o  \]

Thus, the answer is \(0^o \leq x < 45^o\) and \(135^o < x \leq 180^o\).

(2) Use the double angle formula, \(\cos 2x = 1-2\sin^2 x\) to rearrange the given expression.

\[  1-2\sin^2 x +\sin x < 0 \]

\[  2\sin^2 x -\sin x-1 > 0 \]

\[  (2\sin x + 1)(\sin x -1) > 0  \]

\(\sin x -1\) 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

\[  210^o < x < 330^o  \]