Question 217322: Find all the angles between 0° to 360° inclusive which satisfy the equation
i) 2cot^2x + 5 = 7cosecx ii) 2cos^2x + 3sinx = 3
Answer by chibisan(131)   (Show Source):
You can put this solution on YOUR website! i)
cot^2x = cosec^2x -1
2cot^2+5=7cosecx
2(cosec^2x-1)+5 = 7cosecx
2cosec^2x -2 + 5 = 7cosecx
2cosec^2x -7cosecx +3 = 0
(2cosecx-1)(cosecx - 3)= 0
cosecx = 1/2
1/sinx = 1/2
sinx = 2
(no solution)
cosecx=3
1/sinx = 3
sinx =1/3
x = 19.5° , 180 - 19.5°
x = 19.5 , 160.5° (ans)
ii)
2cos^2x + 3sinx = 3
2(1-sin^2x) + 3sinx - 3=0
-2sin^2x + 3sinx + 1= 0
0= 2sin^2x - 3sinx + 1
0= (2sinx- 1)(sinx - 1)
sinx = 1/2
x = 30° , 180 - 30°
x = 30° , 150°
sinx =1
x= 90°, 180°-90°
x = 90°
x= 30° , 90° , 150° (ans)
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