Find the number of solutions of ${sin}^{2} x - sin x - 1 = 0$ in $[- 2 π, 2 π]$

Open in App

Solution

${sin}^{2} x - sin x - 1 = 0$
$\Rightarrow$ $sin x = \frac{1 \pm \sqrt{5}}{2}$
$= \frac{1 - \sqrt{5}}{2}$ [ $sin x = \frac{1 + \sqrt{5}}{2} > 1$ not possible]
Hence, $x$ can attain two values in $[0, 2 π]$ and two more values in $[- 2 π, 0)$ . Thus there are four solutions
Ans: 4

Suggest Corrections

Similar questions

{sin}^{2} x - sin x - 1 = 0

[- 2 π, 2 π]

The number of solution of the equation $s i n x + s i n 2 x + s i n 3 x = c o s x + c o s 2 x + c o s 3 x, 0 \leq x \leq 2 π$ is

s i n x + s i n y + s i n z = - 3 f o r 0 \leq x \leq 2 π, 0 \leq y \leq 2 π, 0 \leq z \leq 2 π

2^{cos x} = | sin x |

x ϵ [- 2 π, 2 π]

| sin x | = | cos 3 x |

[- 2 π, 2 π]

Join BYJU'S Learning Program