How to solve $4 {sin}^{2} (\frac{3 x + π}{6}) = 3$ $4sin^2((3x+pi)/6)=3$ ?

Question

How to solve $4 {sin}^{2} (\frac{3 x + π}{6}) = 3$ $4sin^2((3x+pi)/6)=3$ ?

So, $sin (\frac{3 x + π}{6}) = \pm \frac{\sqrt{3}}{2}$ $sin((3x+pi)/6)=+-sqrt3/2$ . Then what?

2 Answers

Impact of this question

2643 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-06-07T19:11:14

$x = (2 k + 1) π$ $x=(2k+1)pi$ v $(2 k - \frac{1}{3}) π$ $(2k-1/3)pi$

Explanation:

We have $sin (2 \frac{π}{3}) = \frac{\sqrt{3}}{2}$ $sin(2pi/3)=sqrt3/2$ (see https://socratic.org/questions/what-is-sin-2pi-3-equal-to)

Therefore
$sin (\frac{3 x + π}{6}) = \pm sin (2 \frac{π}{3})$ $sin((3x+pi)/6)=+-sin(2pi/3)$
$\frac{3 x + π}{6} = (\frac{2}{3} π + 2 k π)$ $(3x+pi)/6=(2/3pi +2kpi)$ v $(- \frac{2}{3} π + 2 k π)$ $(-2/3pi +2kpi)$
$- \frac{2}{3} π$ $-2/3pi$ should have the same sin value as $\frac{4}{3} π$ $4/3pi$
so that we get
$3 x + π = 6 (\frac{2}{3} π + 2 k π)$ $3x+pi=6(2/3pi +2kpi)$ v $6 (\frac{4}{3} π + 2 k π)$ $6(4/3pi +2kpi)$
$3 x = (4 π - π + 12 k π)$ $3x=(4pi-pi+12kpi)$ v $(12 π - π + 12 k π)$ $(12pi-pi+12kpi)$
$3 x = (3 π + 12 k π)$ $3x=(3pi+12kpi)$ v $(11 π + 12 k π)$ $(11pi+12kpi)$
$x = (π + 4 k π)$ $x=(pi+4kpi)$ v $(\frac{11}{3} π + 4 k π)$ $(11/3pi+4kpi)$
$x = (2 k + 1) π$ $x=(2k+1)pi$ v $(2 k - \frac{1}{3}) π$ $(2k-1/3)pi$

Answer 2 · 2018-06-08T03:51:16

$x = (2 k + 1) \frac{π}{3}$ $x = (2k + 1)pi/3$

Explanation:

$4 {sin}^{2} (\frac{3 x + π}{6}) = 3$ $4sin^2 ((3x + pi)/6) = 3$
$sin (\frac{3 x + π}{6}) = \pm \frac{\sqrt{3}}{2}$ $sin ((3x + pi)/6) = +- sqrt3/2$
a. $sin (\frac{3 x + π}{6}) = \frac{\sqrt{3}}{2}$ $sin ((3x + pi)/6) = sqrt3/2$
Trig table and unit circle give 2 solutions for $\frac{3 x + π}{6}$ $(3x + pi)/6$ :
1. $\frac{3 x + π}{6} = \frac{π}{3}$ $(3x + pi)/6 = pi/3$ --> $3 x + π = 2 π$ $3x + pi = 2pi$
$3 x = 2 π - π = π + 2 k π$ $3x = 2pi - pi = pi + 2kpi$ --> $x = \frac{π}{3} + \frac{2 k π}{3}$ $x = pi/3 + (2kpi)/3$
2. $\frac{3 x + π}{6} = \frac{2 π}{3}$ $(3x + pi)/6 = (2pi)/3$ --> $3 x + π = 4 π$ $3x + pi = 4pi$
$3 x = 3 π + 2 k π$ $3x = 3pi + 2kpi$ --> $x = π + \frac{2 k π}{3}$ $x = pi + (2kpi)/3$
General answer: $x = (2 k + 1) \frac{π}{3}$ $x = (2k + 1)pi/3$
b. $sin (\frac{3 x + π}{6}) = - \frac{\sqrt{3}}{2}$ $sin ((3x + pi)/6) = - sqrt3/2$
Trig table and unit circle give 2 solutions for $\frac{3 x + π}{6}$ $(3x + pi)/6$
1. $\frac{3 x + π}{6} = (4 \frac{π}{3})$ $(3x + pi)/ 6 = (4pi/3)$ --> $(3 x + π) = 8 π$ $(3x + pi) = 8pi$
$3 x = 7 π + 2 k π$ $3x = 7pi + 2kpi$ --> $x = \frac{7 π}{3} + \frac{2 k π}{3}$ $x = (7pi)/3 + (2kpi)/3$ , or
$x = \frac{π}{3} + \frac{2 k π}{3}$ $x = pi/3 + (2kpi)/3$
2. $\frac{3 x + π}{6} = \frac{5 π}{3}$ $(3x + pi)/6 = (5pi)/3$ --> $3 x + π = 10 π$ $3x + pi = 10pi$
$3 x = 9 π + 2 k π$ $3x = 9pi + 2kpi$ --> $x = 3 π + \frac{2 k π}{3}$ $x = 3pi + (2kpi)/3$ , or
x = pi + (2kpi)/3
General answer: $x = (2 k + 1) \frac{π}{3}$ $x = (2k + 1)pi/3$

Science

Math

Humanities

... and beyond

How to solve $4 {sin}^{2} (\frac{3 x + π}{6}) = 3$ $4sin^2((3x+pi)/6)=3$ ?

So, $sin (\frac{3 x + π}{6}) = \pm \frac{\sqrt{3}}{2}$ $sin((3x+pi)/6)=+-sqrt3/2$ . Then what?

2 Answers

Explanation:

Explanation:

Related questions

Impact of this question

Science

Math

Humanities

... and beyond

How to solve 4sin^2((3x+pi)/6)=34sin2(3x+π6)=34sin^2((3x+pi)/6)=3?

So, sin((3x+pi)/6)=+-sqrt3/2sin(3x+π6)=±√32sin((3x+pi)/6)=+-sqrt3/2. Then what?

2 Answers

Explanation:

Explanation:

Related questions

Impact of this question

How to solve $4 {sin}^{2} (\frac{3 x + π}{6}) = 3$ $4sin^2((3x+pi)/6)=3$ ?

So, $sin (\frac{3 x + π}{6}) = \pm \frac{\sqrt{3}}{2}$ $sin((3x+pi)/6)=+-sqrt3/2$ . Then what?