How do you solve $3 cos 2 θ = 2 - sin θ$ $3cos2theta=2-sintheta$ in the interval $0 \leq x \leq 2 π$ $0<=x<=2pi$ ?

Question

How do you solve $3 cos 2 θ = 2 - sin θ$ $3cos2theta=2-sintheta$ in the interval $0 \leq x \leq 2 π$ $0<=x<=2pi$ ?

1 Answer

Impact of this question

3914 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-07-25T03:45:15

$θ = 30^{o}, 150^{o}, {199.47}^{o}$ $theta=30^o,150^o,199.47^o$ or ${340.53}^{o}$ $340.53^o$

Explanation:

$3 cos 2 θ = 2 - sin θ$ $3cos2theta=2-sintheta$ can be simplified as

$3 (1 - 2 {sin}^{2} θ) = 2 - sin θ$ $3(1-2sin^2theta)=2-sintheta$ or

$3 - 6 {sin}^{2} θ = 2 - sin θ$ $3-6sin^2theta=2-sintheta$

$6 {sin}^{2} θ - sin θ - 1 = 0$ $6sin^2theta-sintheta-1=0$

Hence $sin θ = \frac{1 \pm \sqrt{1^{2} - 4 \times 6 \times (- 1)}}{2 \times 6}$ $sintheta=(1+-sqrt(1^2-4xx6xx(-1)))/(2xx6)$

= $\frac{1 \pm \sqrt{1 + 24}}{12} = \frac{1 \pm 5}{12}$ $(1+-sqrt(1+24))/12=(1+-5)/12$

Hence either $sin θ = \frac{1}{2}$ $sintheta=1/2$ i.e. $θ = \frac{π}{6} = 30^{o}$ $theta=pi/6=30^o$ or $\frac{5 π}{6} = 150^{o}$ $(5pi)/6=150^o$

or $sin θ = - \frac{1}{3}$ $sintheta=-1/3$

now as ${sin 19.47}^{o} = \frac{1}{3}$ $sin19.47^o=1/3$ ,

$θ = 180^{\oplus} {19.47}^{o} = {199.47}^{o}$ $theta=180^o+19.47^o=199.47^o$ or $θ = 360^{o} - {19.47}^{o} = {340.53}^{o}$ $theta=360^o-19.47^o=340.53^o$

Science

Math

Humanities

... and beyond

How do you solve $3 cos 2 θ = 2 - sin θ$ $3cos2theta=2-sintheta$ in the interval $0 \leq x \leq 2 π$ $0<=x<=2pi$ ?

1 Answer

Explanation:

Related questions

Impact of this question

Science

Math

Humanities

... and beyond

How do you solve 3cos2theta=2-sintheta3cos2θ=2−sinθ3cos2theta=2-sintheta in the interval 0<=x<=2pi0≤x≤2π0<=x<=2pi?

1 Answer

Explanation:

Related questions

Impact of this question

How do you solve $3 cos 2 θ = 2 - sin θ$ $3cos2theta=2-sintheta$ in the interval $0 \leq x \leq 2 π$ $0<=x<=2pi$ ?