MHB How to Solve Trigonometric Equation sin2x = sqrt(2)/2 Using Algebra?

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To solve the equation sin(2x) = sqrt(2)/2 within the interval 0 to 2π, the initial solutions found are π/8 and 3π/8. The discussion clarifies that to find additional solutions, one can consider the periodic nature of the sine function. By adding 2π to the angles π/4 and 3π/4, the new solutions of 9π/8 and 11π/8 are derived. Dividing these by 2 confirms they fit within the specified interval. The approach effectively utilizes algebraic manipulation and the properties of trigonometric functions.
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Solve sin2x= sqrt(2)/2 (using algebra)
the interval is between 0 and 2pi

i got the first two answers: pi/8 and 3pi/8, but i don t understand how to get the other two: 9pi/8 and 11pi/8.

thanks!
 
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If we are told:

$$0\le x<2\pi$$

then:

$$0\le 2x<4\pi$$
 
Oh, that makes sense!

So, pi/4 + 2pi = 9pi/4
and 3pi/4 + 2pi = 11pi/4

and when i divide them by 2, i get 9pi/8 and 11pi/8.

thank you so much! :D
 
manish00333 said:
So, pi/4 + 2pi = 9pi/4
and 3pi/4 + 2pi = 11pi/4

and when i divide them by 2, i get 9pi/8 and 11pi/8.

Perfect! (Yes)
 
Here is a little puzzle from the book 100 Geometric Games by Pierre Berloquin. The side of a small square is one meter long and the side of a larger square one and a half meters long. One vertex of the large square is at the center of the small square. The side of the large square cuts two sides of the small square into one- third parts and two-thirds parts. What is the area where the squares overlap?

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