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```        FIND ALL ANGLES LYING BETWEEN 0DEGREE AND 360DEGREE SATISFYING
1) 2cos2A-sinA-1=0
2) 4cot2A=3cosec2A
3)sin4A-2sin2A=1
4)3(sec2A+tan2a)=5```
7 years ago

68 Points
```										Dear  kiran ,
1) 2(1- sin2A) -sinA -1 =0
2(1-sin2A))  = sinA + 1
cancelling we get , sinA +1  = 0 ---not possible or 1 - sin2A =1/2 i.e sin2A = 1/2
so , sinA = ±(1/2)0.5   , we get A= 45, 135 , 225 , 315
2) 4 cot2A = 3 cosec2A
4 cos2A =  3  ;  cosA =±  30.5/ 2   , we get A =  30  , 150 , 210 ,330
3)put sin2A = t ,  t2 -2t -1 = 0  solving we get , t  = 1± √2     ; both are not possible . so no solution
4) 3( 1+ sin2A)  = 5 cos2A
3( 1+ sin2A)  = 5 (1- sin2A)
8sin2A  = 2
sin2A  = 1/4
sinA = ± 1/2
we get A  =  30  , 150 , 210 ,330

All the best.
Win exciting gifts by answering the questions on Discussion Forum. So help discuss any query on askiitians forum and become an Elite Expert League askiitian.
Ajit Singh Verma IITD
```
7 years ago
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