
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
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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 ().
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 ().
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