Question

find the number of solutions of sin^2x -sinx - 1 = 0

Answer 1

Answer:

sin2x−sinx−1=0

The above equation is quadratic in sinx

∴sinx=2.1−(−1)±(−1)2−4.1.(−1)

​​⇒sinx=21+5

​​and21−5

​​Here,sin​=21+5

​​(∵21+5

​​>1)∴sinx=21−5

​​

Now, 21−5

​​ is a negative value

If we observe the value of sinx takes all negative values less than −1 at two points

∴sinx=21−5

​​attwopintsin[0,2π]also,sinx=21−5

​​attwopointsin[−2π,0]∴sin2x−sinx−1=0has4solutionsin[−2π,2π]

Answer 2

Answer:

x=1/2

Step-by-step explanation:

2x=0+1

2x=1

x=1/2