If $2 {cos}^{2} x + 7 sin x = 5$ , find the permissible values of $sin x$ .

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Solution

Given,
$2 {cos}^{2} x + 7 sin x = 5$
or, $2 (1 - {sin}^{2} x) + 7 sin x = 5$
or, $2 {sin}^{2} x - 7 sin x + 3 = 0$
or, $2 {sin}^{2} x - 6 sin x - sin x + 3 = 0$
or, $2 sin x (sin x - 3) - 1 (sin x - 3) = 0$
or, $(2 sin x - 1) (sin x + 3) = 0$
As $sin x \neq - 3$ [ Since $- 1 \leq sin x \leq 1$ ]. so $sin x = \frac{1}{2}$ is the required permissible value.

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If 2cos2x+7sinx=5, find the permissible values of sinx.

Given,2cos2x+7sinx=5or, 2(1−sin2x)+7sinx=5or, 2sin2x−7sinx+3=0or, 2sin2x−6sinx−sinx+3=0or, 2sinx(sinx−3)−1(sinx−3)=0or, (2sinx−1)(sinx+3)=0As sinx≠−3[ Since −1≤sinx≤1]. so sinx=12 is the required permissible value.

If $2 {cos}^{2} x + 7 sin x = 5$ , find the permissible values of $sin x$ .