How do you solve $cos 2 x + 5 = 4 sin x$ $cos2x+5=4sinx$ for $0 \leq x \leq 360$ $0<=x<=360$ ?

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How do you solve $cos 2 x + 5 = 4 sin x$ $cos2x+5=4sinx$ for $0 \leq x \leq 360$ $0<=x<=360$ ?

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Answer 1 · 2016-10-25T13:36:36

$\frac{π}{2}$ $pi/2$

Explanation:

cos 2x + 5 = 4sin x
Use trig identity: $cos 2 x = 1 - 2 {sin}^{2} x$ $cos 2x = 1 - 2sin^2 x$
$1 - 2 {sin}^{2} x + 5 = 4 sin x$ $1 - 2sin^2 x + 5 = 4sin x$
Bring the equation to standard form:
$2 {sin}^{2} x + 4 sin x - 6 = 0$ $2sin^2 x + 4sin x - 6 = 0$
Solve this quadratic equation for sin x.
Since a + b + c = 0, use shortcut. There are 2 real roots:
sin x = 1, and
$sin x = \frac{c}{a} = - \frac{6}{2} = - 3$ $sin x = c/a = -6/2 = - 3$ (rejected as < -1)
Unit circle gives -->
sin x = 1 --> $x = \frac{π}{2}$ $x = pi/2$
Answers for $(0, 2 π)$ $(0, 2pi)$ :
$\frac{π}{2}$ $pi/2$
Check.
$x = \frac{π}{2}$ $x = pi/2$ --> sin x = 1 --> $cos 2 x = cos π = - 1$ $cos 2x = cos pi = - 1$
cos 2x + 5 = 4sin x --> - 1 + 5 = 4. OK

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How do you solve $cos 2 x + 5 = 4 sin x$ $cos2x+5=4sinx$ for $0 \leq x \leq 360$ $0<=x<=360$ ?

1 Answer

Explanation:

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How do you solve cos2x+5=4sinxcos2x+5=4sinxcos2x+5=4sinx for 0<=x<=3600≤x≤3600<=x<=360?

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How do you solve $cos 2 x + 5 = 4 sin x$ $cos2x+5=4sinx$ for $0 \leq x \leq 360$ $0<=x<=360$ ?