Materi : ID . 3GONOMET3 Kelas : XI SMA Dijawab ya kak/om/bu' ....

23)
Diketahui:
*tetha nya gak ada di keyboard jadi pake simbol @ aja yaa.*
0° < @ < 90°

1 + sin 2@ = 2 cos² 2@
sin 2@ = 2 cos² 2@ - 1
sin 2@ = cos (2@+2@)
*rumus cos sudut rangkap*
sin 2@ = cos 4@

sin 2@ = sin (90°-4@)
2@ = (90° - 4@) ± k.360°
6@ = 90° ± k.360°
@ = 15° ± k.60°
@ = 15° dan @ = 75°

sin 2@ = sin (90°-4@)
sin 2@ = sin (180°-(90°-4@))
2@ = (90° + 4@) ± k.360°
-2@ = 90° ± k.360°
@ = -45° ± k.180°
*tidak memenuhi*


1 + sin 2@ = 2 cos² 2@
1 + sin 2@ - 2(1 - sin² 2@) = 0
2 sin² 2@ + sin 2@ - 1 = 0
(2 sin 2@ - 1)(sin 2@ + 1) = 0

sin 2@ = 1/2
sin 2@ = sin 30°
@ = 15°
dan
sin 2@ = -1
sin 2@ = sin 270°
@ = 135°
*tidak memenuhi*

Jadi @ yang memenuhi {15°, 75°} C



24)
180° < x < 270°

9 tan² x - 24 tan x + 16 = 0
(3 tan x - 4)² = 0
3 tan x = 4
tan x = 4/3

Karena x di Kuadran III maka
sin x = [tex] \frac{-4}{\sqrt{(-4)^2+(-3)^2}} [/tex]
sin x = -[tex] \frac{4}{5} [/tex]
Jawabannya D


25)

12 sin² x + sin x - 1 = 0
*menggunakan rumus abc*
sin x1 = 1/4
dan
sin x2 = -1/3

csc² x1 + csc² x2 - 7
= (4)² + (-3)² - 7
= 16 + 9 - 7
= 18
Jawabannya A