The set of all x in the interval -0 - - for which sin2 x -3sin x - 1 - 0 is

f(x) $= \frac{π}{4} + \frac{2}{π} [\frac{cos x}{1^{2}} + \frac{cos 3 x}{3^{2}} + . . . .] + [\frac{sin x}{1} + \frac{sin 2 x}{2} + \frac{sin 3 x}{3} + . . . .]$ The convergence of the above Fourier series at x = 0 gives

Q. The intervals for which

tan x > cot x

, where

x \in (0, π) - {\frac{π}{2}}

Join BYJU'S Learning Program

The set of all x in the interval [0,π] for which sin2x−3sinx+1≥0 is

The correct option is D none of thesesin2x−3sinx+1≥0⟹sinx=3±√9−42⟹sinx=3±√52since the range of sin function in [0,π] is (0,1)So there is no solution.

The set of all x in the interval $[0, π]$ for which ${sin}^{2} x - 3 sin x + 1 \geq 0$ is

The correct option is D none of these
$s i n^{2} x - 3 s i n x + 1 \geq 0$

$⟹ s i n x = \frac{3 \pm \sqrt{9 - 4}}{2}$

$⟹ s i n x = \frac{3 \pm \sqrt{5}}{2}$

since the range of sin function in $[0, π]$ is (0,1)

So there is no solution.