Question #9b19b

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Answer 1 · 2018-01-07T05:29:05

$x = {sin}^{- 1} (\frac{\sqrt{5}}{5})$ $x=sin^-1(sqrt5 /5 )$

Assuming your problem is: $4 {cos}^{2} x = {sin}^{2} x + 3$ $4cos^2x=sin^2x+3$

Using ${sin}^{2} x + {cos}^{2} x = 1$ $sin^2x+cos^2x=1$ , we get

$4 (1 - {sin}^{2} x) = {sin}^{2} x + 3$ $4(1-sin^2x)=sin^2x+3$

$4 - 4 {sin}^{2} x = {sin}^{2} x + 3$ $4-4sin^2x=sin^2x+3$

$5 {sin}^{2} x = 1$ $5sin^2x=1$

${sin}^{2} x = 0.2$ $sin^2x=0.2$

$\Rightarrow sin x = \sqrt{0.2} = \frac{\sqrt{5}}{5}$ $=> sinx = sqrt0.2 = sqrt5 / 5$

$x = {sin}^{- 1} (\frac{\sqrt{5}}{5}) \approx 0.4647$ $x=color(red)(sin^-1(sqrt5 / 5 )~~0.4647$ $r a d$ $color(red)(rad$