Solve the equation for solutions in the interval [0, 2).sin22x = 1O O프O97{}8 8}37 57○꾸쭈꾸}

We have given trigonometric equation sin22x=1 sin2x=1 sin2x=±1 i sin2x=1 sin2x=sinπ2=sin2π+π2…

Answered: Solve the equation for solutions in the…

Trigonometry (11th Edition)

11th Edition

ISBN:9780134217437

Author:Margaret L. Lial, John Hornsby, David I. Schneider, Callie Daniels

Publisher:Margaret L. Lial, John Hornsby, David I. Schneider, Callie Daniels

Chapter1: Trigonometric Functions

Section: Chapter Questions

Problem 1RE: 1. Give the measures of the complement and the supplement of an angle measuring 35°.

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Question

Solve the equation for solutions in the interval [0, 2).
sin22x = 1
O O
프
O
97
{}
8 8
}
37 57
○꾸쭈꾸}

Expert Solution

Step 1

We have given trigonometric equation

$\sin^{2} 2 x = 1$

$\sin 2 x = |1|$

$\sin 2 x = \pm 1$

$(i)$ $\sin 2 x = 1$

$\sin 2 x = \sin$ $(\frac{π}{2}) = \sin (2 π + \frac{π}{2})$

$\sin 2 x = \sin$ $(\frac{π}{2}) = \sin (\frac{5 π}{2})$

$2 x = \frac{π}{2}, \frac{5 π}{2}$

$x = \frac{π}{4}, \frac{5 π}{4}$

(ii) $\sin 2 x = - 1$

$\sin 2 x = -$ $\sin (\frac{π}{2})$

$\sin 2 x =$ $\sin (π + \frac{π}{2}) =$ $= \sin (3 π + \frac{π}{2})$

$\sin 2 x = \sin$ $(\frac{3 π}{2}) = \sin$ $(\frac{7 π}{2})$

$2 x = \frac{3 π}{2}, \frac{7 π}{2}$

$x = \frac{3 π}{4}, \frac{7 π}{4}$

So, we get

x = $\frac{π}{4}, \frac{3 π}{4}, \frac{5 π}{4},$ $\frac{7 π}{4}$

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Solve the equation for solutions in the interval [0, 2n). sin22x = 1 {끝 블} 97 ○ {풍증} 8 37 57 ○꾸쭈꾸