How do you solve sin2x = √2 cosx, 0≤x≤ π?

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How do you solve sin2x = √2 cosx, 0≤x≤ π?

sin2x = $\sqrt{2}$ $sqrt(2)$ cosx, 0 $\leq$ $<=$ x $\leq$ $<=$ $π$ $pi$ ?

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Answer 1 · 2017-03-24T16:28:18

$\frac{π}{2}; \frac{π}{4}; \frac{3 π}{4}$ $pi/2; pi/4; (3pi)/4$

Explanation:

Use trig identity : sin 2x = 2sin x.cos x
Substitute in the equation sin 2x by 2sin x.cos x
$2 sin x . cos x - \sqrt{2} cos x = 0$ $2sin x.cos x - sqrt2cos x = 0$
$cos x (2 sin x - \sqrt{2}) = 0$ $cos x(2sin x - sqrt2) = 0$
Either one of the 2 factors should be zero:
a. cos x = 0 --> $x = \frac{π}{2}$ $x = pi/2$ , and $x = \frac{3 π}{2}$ $x = (3pi)/2$
b. $2 sin x - \sqrt{2} = 0$ $2sin x - sqrt2 = 0$
$sin x = \frac{\sqrt{2}}{2}$ $sin x = sqrt2/2$
Trig unit circle gives 2 solutions:
$x = \frac{π}{4} and x = \frac{3 π}{4}$ $x = pi/4 and x = (3pi)/4$
Answers for $(0, π)$ $(0, pi)$
$\frac{π}{2}; \frac{π}{4}; \frac{3 π}{4}$ $pi/2; pi/4; (3pi)/4$

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How do you solve sin2x = √2 cosx, 0≤x≤ π?

sin2x = $\sqrt{2}$ $sqrt(2)$ cosx, 0 $\leq$ $<=$ x $\leq$ $<=$ $π$ $pi$ ?

1 Answer

Explanation:

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Math

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How do you solve sin2x = √2 cosx, 0≤x≤ π?

sin2x = sqrt(2)√2sqrt(2) cosx, 0<=≤<=x<=≤<=piπpi?

1 Answer

Explanation:

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sin2x = $\sqrt{2}$ $sqrt(2)$ cosx, 0 $\leq$ $<=$ x $\leq$ $<=$ $π$ $pi$ ?