Answer:
E. 45° + 360°n, 135° + 360°n, 225° + 360°n, 315° + 360°n
Step-by-step explanation:
Given trigonometric equation:
[tex]\sin^2(2x)=1[/tex]
To solve the given trigonometric equation for all x, begin by taking the square root of both sides:
[tex]\sin(2x)=\pm 1[/tex]
According to the unit circle, the sine of an angle is equal to 1 when θ = 90°, and the sine of an angle is equal to -1 when θ = 270°. Therefore:
[tex]2x=90^{\circ} \\\\2x=270^{\circ}[/tex]
Since the sine function is a periodic with a period of 360°, it repeats its values every 360°. Therefore, when solving trigonometric equations involving sine, we need to add multiples of 360° to each solution to account for all possible solutions within the given range. Therefore:
[tex]2x=90^{\circ}+360^{\circ}n \\\\2x=270^{\circ}+360^{\circ}n[/tex]
To solve for x, divide both sides of each equation by 2:
[tex]\dfrac{2x}{2}=\dfrac{90^{\circ}+360^{\circ}n}{2}\\\\\\x=45^{\circ}+180^{\circ}n[/tex] [tex]\dfrac{2x}{2}=\dfrac{270^{\circ}+360^{\circ}n}{2}\\\\\\x=135^{\circ}+180^{\circ}n[/tex]
So, the solutions of the given trigonometric equation, where n is an integer, are:
[tex]x=45^{\circ}+180^{\circ}n\\\\x=135^{\circ}+180^{\circ}n[/tex]
To rewrite this using +360°n notation, find all the solutions within the interval 0 ≤ x ≤ 360°, then add 360°n to each. Therefore, the solutions are:
[tex]\Large\boxed{\begin{array}{l}x=45^{\circ}+360^{\circ}n\\x=135^{\circ}+360^{\circ}n\\x=225^{\circ}+360^{\circ}n\\x=315^{\circ}+360^{\circ}n\end{array}}[/tex]
