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)
 
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