Question

16. (i) 4 cos2 x – 4 sin x – 1 = 0 find general solution​

Answer 1

Answer:

4cos²x - 4sinx - 1 = 0  

=> 4(1-sin²x) - 4sinx - 1 = 0 =>

4 - 4sin²x - 4sinx - 1 = 0 =>    

4sin²x - 4sinx + 3 = 0 =>

4sin²x + 4sinx - 3 = 0 =>

4sin²x + 6sinx - 2sinx - 3 = 0 =>

2sinx(2sinx+3) - 1(2sinx+3) = 0 =>

(2sinx-1)(2sinx+3) = 0 =>

(2sinx-1) = 0  or   (2sinx+3) = 0 =>

sinx = 1/2      Or   sinx = -3/2 =>

sinx = sin30° --sinx cannot be equal to -3/2 because value of sin lies between -1 and 1  

So x = 30°

Step-by-step explanation:

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