How do you find all solutions of the equation in the interval $[0, 2 π)$ $[0,2pi)$ given ${sin}^{2} x + 2 cos x = 2$ $sin^2x+2cosx=2$ ?

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How do you find all solutions of the equation in the interval $[0, 2 π)$ $[0,2pi)$ given ${sin}^{2} x + 2 cos x = 2$ $sin^2x+2cosx=2$ ?

1 Answer

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Answer 1 · 2018-06-18T06:44:13

x=0

Explanation:

${sin}^{2} x = 1 - {cos}^{2} x$ $sin^2 x=1-cos^2 x$
$1 - {cos}^{2} x + 2 cos x = 2$ $1-cos^2 x + 2 cos x = 2$
let $A = cos x$ $A=cosx$
Now solve
$- A^{2} + 2 A - 1 = 0$ $-A^2+2A-1=0$
$- {(A - 1)}^{2} = 0$ $-(A-1)^2=0$
$A = cos x = 1$ $A=cosx=1$
$x = 2 n π$ $x=2npi$ (n=integer number)
In the interval $[0, 2 π)$ $[0,2pi)$
the only solution is
when n=0 x=0

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How do you find all solutions of the equation in the interval $[0, 2 π)$ $[0,2pi)$ given ${sin}^{2} x + 2 cos x = 2$ $sin^2x+2cosx=2$ ?

1 Answer

Explanation:

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How do you find all solutions of the equation in the interval [0,2pi)[0,2π)[0,2pi) given sin^2x+2cosx=2sin2x+2cosx=2sin^2x+2cosx=2?

1 Answer

Explanation:

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How do you find all solutions of the equation in the interval $[0, 2 π)$ $[0,2pi)$ given ${sin}^{2} x + 2 cos x = 2$ $sin^2x+2cosx=2$ ?