Solve each equation on the interval:
0 is less than and equal to theta is less than and equal to 360
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2sin2x = 1
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My Answers:
Let y represent 2x
2siny = 1
2siny/2 = 1/2
siny = 1/2
y = sin^-1(1/2)
y = 30
2x = 30
2x/2 = 30/2
x = 15 (Quadrant 1)
Quadrant 2
180 - theta
= 180 - 30
= 150
2x = 150
2x/2 = 150/2
x = 75
Therefore, x = 15 and 75 degrees
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Book's Answers:
15, 75, 195, 255
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My Problem:
I'm asking how do you get 195 and 255 degrees if the equation is sin and it's positive and the restriction is between 0 and 360 degrees, so angles are on Quadrant 1 and 2?
This is how I would do it. (I've never used y to represent 2x, but maybe your teacher does that.)
2sin2x = 1
sin2x = 1/2
If we reference the unit circle (which should be in your head), then we know that sine corresponds to the y-coordinate. y=1/2 happens at 30 and 150 degrees. Since we have "2x," we must make two passes around the unit circle, so add 360 to both of those angles. 30+360 = 390 and 150+360 = 510.
Therefore, 2x = 30, 150, 390, 510.
Divide all the angles by 2 to isolate x.
x = 15, 75, 195, 255
You did a good job, but you just forgot to make the second pass. (The angles are above 360, but when we divide by 2, they are within our restriction.)
Here is a slightly different way of looking at those extra angles than your text book gave you.
Everything is fine until you found x = 15 and 75, even though, as "anonymous" also suggested, there is really no need to make the substitution y = 2x.
Look back at the actual function containing sin2x, the period of sin kx (or cos kx for that matter) is 360º/k
so for your function the period is 360/2 = 180º
That means that every answer you obtained will repeat itself 180º in either the positive or negative direction.
so simply add 180º or subtract 180º, to each of the answers that you already have
This way there is really an infinite number of solutions to your equation, others being 195, 375º, 555º,...
255,435,...-165,-345,...etc
All you have to do is choose those that lie in the given domain
Reiny, I was anonymous... my name wasn't filled in. ;)
To clarify, the equation you are solving is 2sin(2x) = 1 for the interval 0 ≤ θ ≤ 360.
You've correctly found the solutions in Quadrants 1 and 2, which are x = 15° and x = 75°.
To find the other solutions within the given interval, you need to consider the periodic nature of the sine function. Since the period of sin(θ) is 360°, you can add/subtract multiples of 360° to find additional solutions.
Let's start with x = 15° and find the next solution:
x = 15° + 360° = 375°
Now let's find the next solution using x = 75°:
x = 75° + 360° = 435°
However, these solutions are not within the given interval of 0 ≤ θ ≤ 360. To bring them back into the interval, you need to subtract 360° from them:
375° - 360° = 15°
435° - 360° = 75°
Therefore, the correct solutions within the interval 0 ≤ θ ≤ 360 are x = 15°, x = 75°, x = 15° + 180° = 195°, and x = 75° + 180° = 255°.