Question #0967f

Question

Question #0967f

2 Answers

Related questions

How do you find all solutions trigonometric equations?
How do you express trigonometric expressions in simplest form?
How do you solve trigonometric equations by factoring?
How do you solve trigonometric equations by the quadratic formula?
How do you use the fundamental identities to solve trigonometric equations?
What are other methods for solving equations that can be adapted to solving trigonometric equations?
How do you solve ${sin}^{2} x - 2 sin x - 3 = 0$ $\sin^2 x - 2 \sin x - 3 = 0$ over the interval $[0, 2 π]$ $[0,2pi]$ ?
How do you find all the solutions for $2 {sin}^{2} \frac{x}{4} - 3 cos \frac{x}{4} = 0$ $2 \sin^2 \frac{x}{4}-3 \cos \frac{x}{4} = 0$ over the...
How do you solve ${cos}^{2} x = \frac{1}{16}$ $\cos^2 x = \frac{1}{16}$ over the interval $[0, 2 π]$ $[0,2pi]$ ?
How do you solve for x in $3 sin 2 x = cos 2 x$ $3sin2x=cos2x$ for the interval $0 ≤ x < 2π$

Impact of this question

1989 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-10-01T19:40:18

X = 0.201(approx.)

Explanation:

${sin (sin x)}^{2} = \frac{1}{25}$ $sin(sinx)^2 = 1/25$
${sin}^{2} x = {sin}^{- 1} (\frac{1}{25})$ $sin^2 x = sin^-1(1/25)$
$sin x = \pm \sqrt{{sin}^{- 1} (\frac{1}{25})}$ $sinx = ±sqrt {sin^-1(1/25)}$
${sin}^{- 1} [\pm \sqrt{{sin}^{- 1} (\frac{1}{25})}] = x$ $sin^-1 [±sqrt {sin^-1(1/25)} ] = x$

Answer 2 · 2016-10-03T11:19:50

$sin (x^{2}) = \frac{1}{25}$ $sin(x^2) =1/25$
$x^{2} = {sin}^{- 1} (\frac{1}{25})$ $x^2=sin^-1 (1/25)$
$x = \pm \sqrt{{sin}^{- 1} (\frac{1}{25})}$ $x=±sqrt (sin^-1 (1/25))$ = y let
Now $x = y, π - y, π + y, 2 π - y$ $x= y,pi-y,pi+y,2pi-y$

Science

Math

Humanities

... and beyond

Question #0967f

2 Answers

Explanation:

Related questions

Impact of this question