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

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 =±  3^0.5/ 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

 Please feel free to ask your queries here. We are all IITians and here to help you in your IIT JEE preparation.

All the best.

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