How do you find the exact solutions of the equation ${(sin 2 x + cos 2 x)}^{2} = 1$ $(sin2x+cos2x)^2=1$ in the interval $[0, 2 π)$ $[0,2pi)$ ?

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How do you find the exact solutions of the equation ${(sin 2 x + cos 2 x)}^{2} = 1$ $(sin2x+cos2x)^2=1$ in the interval $[0, 2 π)$ $[0,2pi)$ ?

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Answer 1 · 2017-01-25T14:37:32

Exact solution in interval $[0, 2 π)$ $[0,2pi)$ is ${0, \frac{π}{4}, \frac{π}{2}, \frac{3 π}{4}, π, \frac{5 π}{4}, \frac{3 π}{2}, \frac{7 π}{4}}$ ${0,pi/4,pi/2,(3pi)/4,pi,(5pi)/4,(3pi)/2,(7pi)/4}$

Explanation:

${(sin 2 x + cos 2 x)}^{2} = 1$ $(sin2x+cos2x)^2=1$ can be expanded as

${sin}^{2} (2 x) + {cos}^{2} (2 x) + 2 sin 2 x cos 2 x = 1$ $sin^2(2x)+cos^2(2x)+2sin2xcos2x=1$

or ${sin}^{2} (2 x) + {cos}^{2} (2 x) + 2 sin 2 x cos 2 x = 1$ $sin^2(2x)+cos^2(2x)+2sin2xcos2x=1$

or $1 + sin 4 x = 1$ $1+sin4x=1$

i.e. $sin 4 x = 0$ $sin4x=0$

and hence $4 x = n π$ $4x=npi$ and $x = \frac{n π}{4}$ $x=(npi)/4$

Hence, exact solution in interval $[0, 2 π)$ $[0,2pi)$ is ${0, \frac{π}{4}, \frac{π}{2}, \frac{3 π}{4}, π, \frac{5 π}{4}, \frac{3 π}{2}, \frac{7 π}{4}}$ ${0,pi/4,pi/2,(3pi)/4,pi,(5pi)/4,(3pi)/2,(7pi)/4}$

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How do you find the exact solutions of the equation ${(sin 2 x + cos 2 x)}^{2} = 1$ $(sin2x+cos2x)^2=1$ in the interval $[0, 2 π)$ $[0,2pi)$ ?

1 Answer

Explanation:

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Science

Math

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How do you find the exact solutions of the equation (sin2x+cos2x)2=1(sin2x+cos2x)^2=1 in the interval [0,2π)[0,2pi)?

1 Answer

Explanation:

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How do you find the exact solutions of the equation ${(sin 2 x + cos 2 x)}^{2} = 1$ $(sin2x+cos2x)^2=1$ in the interval $[0, 2 π)$ $[0,2pi)$ ?