Posted by Anonymous on Wednesday, November 21, 2007 at 8:23pm.
Solve each equation on the interval:
0 is less than and equal to theta is less than and equal to 360
2sin2x = 1
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)
180 - theta
= 180 - 30
2x = 150
2x/2 = 150/2
x = 75
Therefore, x = 15 and 75 degrees
15, 75, 195, 255
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?
- Math - Solving Trig Equations - Anonymous, Wednesday, November 21, 2007 at 8:33pm
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.)
- Math - Solving Trig Equations - Reiny, Wednesday, November 21, 2007 at 9:05pm
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,...
All you have to do is choose those that lie in the given domain
Answer This Question
More Related Questions
- Math - Solving Trig Equations - Solve each equation for o is less than and/or ...
- Precalculs - I have no idea how to do these type of problems. -------Problem...
- Math - Solving Trig Equations - Solve each equation for 0 is less than and equal...
- Math - Solving for Trig Equations - Solve the following equation for 0 less than...
- trig - What are the values of theta in the interval 0 less than or equal to ...
- algebra/trigonometry - What are the values of theta in the interval 0degrees ...
- trig - what are the values of theta in the interval 0 degrees is lessthan or ...
- Trig - Given 2secant^2 (theta) + 1 = 5 secant (theta), solve for (theta), to the...
- Math Trig Help - sin theta = x/3, with theta in the first quadrant, find the ...
- Precalc with Trig - What value(s) of theta (0 less than or equal to theta less ...