How do you solve $1 + sin x = 2 {cos}^{2} x$ $1+sinx=2cos^2x$ ?

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How do you solve $1 + sin x = 2 {cos}^{2} x$ $1+sinx=2cos^2x$ ?

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Answer 1 · 2016-11-24T21:36:47

$x = 30^{o} + {(- 1)}^{n} (180^{o} \times n)$ $x = 30^o +(-1)^n(180^o xxn)$ or $x = - 90^{o} + (360^{o} \times n)$ $x =-90^o +(360^o xxn)$

where n can be any positive or negative integer, including 0.

Explanation:

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

Substitute $1 - {sin}^{2} (x)$ $1 - sin^2(x)$ for ${cos}^{2} (x)$ $cos^2(x)$ :

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

Distribute the 2:

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

Add $2 {sin}^{2} (x) - 2$ $2sin^2(x) - 2$ to both sides:

$2 {sin}^{2} (x) + sin (x) - 1 = 0$ $2sin^2(x) + sin(x) - 1 = 0$

Divide by 2:

${sin}^{2} (x) + \frac{1}{2} sin (x) - \frac{1}{2} = 0$ $sin^2(x) + 1/2sin(x) - 1/2 = 0$

Use the quadratic formula:

$sin (x) = \frac{- \frac{1}{2} \pm \sqrt{{(\frac{1}{2})}^{2} - 4 (1) (- \frac{1}{2})}}{2 (1)}$ $sin(x) = (-1/2 +-sqrt((1/2)^2 - 4(1)(-1/2)))/(2(1))$

$sin (x) = \frac{- \frac{1}{2} \pm \sqrt{\frac{1}{4} + 2}}{2}$ $sin(x) = (-1/2 +-sqrt(1/4 + 2))/(2)$

$sin (x) = \frac{- \frac{1}{2} \pm \sqrt{\frac{9}{4}}}{2}$ $sin(x) = (-1/2 +-sqrt(9/4))/(2)$

$sin (x) = \frac{- \frac{1}{2} \pm \frac{3}{2}}{2}$ $sin(x) = (-1/2 +-3/2)/(2)$

$sin (x) = \frac{- 1 + 3}{4} and sin (x) = \frac{- 1 - 3}{4}$ $sin(x) = (-1 + 3)/4 and sin(x)=(-1 - 3)/4$

$x = {sin}^{- 1} (\frac{1}{2}) and {sin}^{- 1} (- 1)$ $x = sin^-1(1/2) and sin^-1(-1)$

I am going to use degrees, because the numbers are much nicer than the radian values.

$x = 30^{o}$ $x = 30^o$ or $150^{o}$ $150^o$ and $x = - 90^{o}$ $x=-90^o$

This will repeat an integer multiples of $360^{o}$ $360^o$

$x = 30^{o} + {(- 1)}^{n} (180^{o} \times n)$ $x = 30^o +(-1)^n(180^o xxn)$ or $x = - 90^{o} + (360^{o} \times n)$ $x =-90^o +(360^o xxn)$

where n can be any positive or negative integer, including 0.

Answer 2 · 2016-11-25T15:49:28

$\frac{π}{2} + 2 k π$ $pi/2 + 2kpi$
$\frac{7 π}{6} + 2 k π$ $(7pi)/6 + 2kpi$
$\frac{11 π}{6} + 2 k π$ $(11pi)/6 + 2kpi$

Explanation:

$1 + sin x = 2 {cos}^{2} x$ $1 + sin x = 2cos^2 x$
Replace $(2 {cos}^{2} x) b y (2 (1 - {sin}^{2} x))$ $(2cos^2 x) by (2( 1 - sin^2 x))$ :
$1 + sin x = 2 (1 - {sin}^{2} x) = 2 - 2 {sin}^{2} x$ $1 + sin x = 2( 1 - sin^2 x) = 2 - 2sin^2 x$
Bring quadratic equation to standard form, and solve it
$2 {sin}^{2} x - sin x - 1 = 0$ $2sin^2 x - sin x - 1 = 0$
Since a + b + c = 0, use shortcut. The 2 real roots are:
sin x = 1 and $sin x = \frac{c}{a} = - \frac{1}{2}$ $sin x = c/a = - 1/2$
a. sin x = 1 --> arc $x = \frac{π}{2}$ $x = pi/2$
b. $sin x = - \frac{1}{2}$ $sin x = - 1/2$
Trig unit circle gives 2 solutions
$sin x = - \frac{1}{2}$ $sin x = -1/2$ --> arc $x = - \frac{π}{6}$ $x = - pi/6$ and arc $x = - \frac{5 π}{6}$ $x = - (5pi)/6$
Solution for $(0, 2 π)$ $(0, 2pi)$
$\frac{π}{2}, \frac{7 π}{6}, \frac{11 π}{6}$ $pi/2, (7pi)/6, (11pi)/6$
The arc $\frac{7 π}{6} and \frac{11 π}{6}$ $(7pi)/6 and (11pi)/6$ are co-terminal to the arcs $\frac{- 5 π}{6} and (- \frac{π}{6})$ $(- 5pi)/6 and (-pi/6)$ .
For general answers, add $2 k π$ $2kpi$ for each solution

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How do you solve $1 + sin x = 2 {cos}^{2} x$ $1+sinx=2cos^2x$ ?

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