Question

The equation \[{\sin ^4}x + {\cos ^4}x + \sin 2x + \alpha = 0\] is solvable for
A. \[ - \frac{1}{2} \le \alpha \le \frac{1}{2}\]
B. \[ - 3 \le \alpha \le 1\]
C. \[ - \frac{3}{2} \le \alpha \le \frac{1}{2}\]
D. \[ - 1 \le \alpha \le 1\]

Answer 1

Hints
The solutions to the trigonometric equation will then be discovered using the provided interval. An equation incorporating the trigonometric ratios is referred to as a trigonometric equation. Here, we must determine alpha's range. Here, the value of \[{\sin ^4}x + {\cos ^4}x + \sin 2x + \alpha = 0\]is provided. By expressing \[\sin 2x\]and\[\cos 2x\]formula in terms of \[\cos 2x\]in the value of a, we will first broaden the terms. We shall translate alpha’s value into \[\cos 2x\]. We'll employ the \[\cos 2x\] range, which lies between\[ - 1\]and\[1\]. We shall obtain the necessary range of alpha from there.
Formula used:
\[{{\rm{a}}^{\rm{2}}}{\rm{ + }}{{\rm{b}}^{{\rm{2 }}}}{\rm{ = (a + b}}{{\rm{)}}^{\rm{2}}}{\rm{ - 2ab}}\]
Complete step-by-step solution
We have been given the equation
\[{\sin ^4}x + {\cos ^4}x + \sin 2x + \alpha = 0\]
Write using square formula
\[{\left( {{{\sin }^2}x} \right)^2} + {\left( {{{\cos }^2}x} \right)^2} + \sin 2x + \alpha = 0\]
Expand the above equation using formula:
\[{\left( {\sin {n^2}x + {{\cos }^2}x} \right)^2} - 2{\sin ^2}x \cdot {\cos ^2}x + \sin 2x + \alpha = 0\]
Simplify the equation to make it less complicated:
\[1 - 2{\sin ^2}x \cdot {\cos ^2}x + \sin 2x + \alpha = 0\]
So, the equation becomes,
\[1 - 2\frac{{{{\sin }^2}2x}}{4} + \sin 2x + \alpha = 0\]
Solve the above equation to make it even simpler form:
\[2 - {\sin ^2}2\alpha + 2\sin 2x + 2\alpha = 0\]
Now, solve for\[2\alpha \]:
\[2\alpha = {\sin ^2}2\alpha - 2\sin 2x - 2\]
Replace the values with \[y\]:
\[2\alpha = {y^2} - 2y - 2\]
Expand using formula:
\[2\alpha = {(y - 1)^2} - 3\]
Now, solve the obtained equation for \[\alpha \]:
\[\alpha = \frac{{{{(y - 1)}^2}}}{2} - \frac{3}{2}\]
The solutions can be obtained as,
\[ - 1 \le y \le 1\] = \[0 \le {(y - 1)^2} \le 4\] = \[ - 2 \le y - 1 \le 0\]
Simplify the above expression to determine the intervals
\[ - 3 \le {(y - 1)^2} - 3 \le 1\] = \[ - 3 \le 2\alpha \le 1\]
Therefore, the equation\[{\sin ^4}x + {\cos ^4}x + \sin 2x + \alpha = 0\]is solvable for\[ - \frac{3}{2} \le \alpha \le \frac{1}{2}\]
Hence, the option C is correct.
Note
We are aware that the trigonometric identity is a mathematical equation that holds true regardless of the variables. The ratios of a right-angled triangle's sides to any of its acute angles are known as trigonometric ratios of a particular angle. We are aware that the missing angles of a right-angled triangle can be found using the inverse trigonometric function. Knowing that an Open interval is a period of time that excludes the endpoints and is indicated by ().