Trigonometric Roots of sin4x = -(1/sqrt[2]) on 0 <= x <= 2pi

Re: Trigonometric Roots

Suppose it said sin A = 1/2.
Can you come up with all the angles between 0° and 360° for which sin A = 1/2? I think there are two, namely A= 60° and A=120°.

What if it said sin 3A = 1/2? You would then say that 3A=60° or 3A=120°. Therefore, A=20° or A=40°.

But, wait. What if it said "for all A between 0° and 360°. We would have to find all the angles between 0° and 3*360=1080° for 3A, because of our division by 3 which will put us back between 0 and 360°.
 
Re: Trigonometric Roots

sail0r said:
Find the roots for sin4x= -(1/?2) 0?x?2?

How is this done?
Click to expand...

Check unit circle and find out for which angle/s - sine of which is - -(1/?2)

Then equate 4x to that and solve.

To get more help - you need to show your work, indicating exactly where you are stuck, so that we know where to begin to help you.
 
When the equation has sin 4x and 0°? x ? 360°, then 0°< 4x ? 1440°. This means that you need to travel around the unit circle 4 times to find all the solutions.
 
fasteddie65 said:
When the equation has sin 4x and 0°? x ? 360°, then 0°< 4x ? 1440° ...
Click to expand...


Of course, this is not what the exercise has, so I will correct.

"When the equation has sin(4x) = -1/?2 and 0 ? x ? 2?, then 0 ? 4x ? 8?."

 
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