Solve $sin x cos$ $sinxcos$ $\frac{π}{6} + cos x sin x$ $pi/6+cosxsinx$ $\frac{π}{6} = \frac{1}{\sqrt{2}}$ $pi/6=1/sqrt2$ for $0 \leq x, = 2 π$ $0<=x,=2pi$ ?

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Solve $sin x cos$ $sinxcos$ $\frac{π}{6} + cos x sin x$ $pi/6+cosxsinx$ $\frac{π}{6} = \frac{1}{\sqrt{2}}$ $pi/6=1/sqrt2$ for $0 \leq x, = 2 π$ $0<=x,=2pi$ ?

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Answer 1 · 2017-05-30T04:37:38

$\frac{π}{12}; \frac{7 π}{12}$ $pi/12; (7pi)/12$

Explanation:

Develop the left side
$L S = sin x . cos (\frac{π}{6}) + cos x . sin (\frac{π}{6}) = sin (x + \frac{π}{6})$ $LS = sin x.cos (pi/6) + cos x.sin (pi/6) = sin (x + pi/6)$
Equation to solve:
$sin (x + \frac{π}{6}) = \frac{1}{\sqrt{2}}$ $sin (x + pi/6) = 1/sqrt2$
Trig table and unit circle give 2 solutions:
a. $x + \frac{π}{6} = \frac{π}{4}$ $x + pi/6 = pi/4$
x = $\frac{π}{4} - \frac{π}{6} = \frac{π}{12}$ $pi/4 - pi/6 = pi/12$
b. $x + \frac{π}{6} = π - \frac{π}{4} = \frac{3 π}{4}$ $x + pi/6 = pi - pi/4 = (3pi)/4$
$x =$ $x =$ (3pi)/4 - pi/6 = (7pi)/12#

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Solve $sin x cos$ $sinxcos$ $\frac{π}{6} + cos x sin x$ $pi/6+cosxsinx$ $\frac{π}{6} = \frac{1}{\sqrt{2}}$ $pi/6=1/sqrt2$ for $0 \leq x, = 2 π$ $0<=x,=2pi$ ?

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Explanation:

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Solve sinxcossinxcossinxcospi/6+cosxsinxπ6+cosxsinxpi/6+cosxsinxpi/6=1/sqrt2π6=1√2pi/6=1/sqrt2 for 0<=x,=2pi0≤x,=2π0<=x,=2pi ?

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Solve $sin x cos$ $sinxcos$ $\frac{π}{6} + cos x sin x$ $pi/6+cosxsinx$ $\frac{π}{6} = \frac{1}{\sqrt{2}}$ $pi/6=1/sqrt2$ for $0 \leq x, = 2 π$ $0<=x,=2pi$ ?