Solve for theta.
4 sin²theta - 2 (√3+1) sin theta + √3 = 0
Answers
Answer:
\theta = \dfrac{\pi }{3}θ=
3
π
and
\theta = \dfrac{\pi }{6}θ=
6
π
Step-by-step explanation:
Given that-
4Sin^{2}\theta -2(\sqrt{3}+1)Sin\theta +\sqrt{3} = 04Sin
2
θ−2(
3
+1)Sinθ+
3
=0
4Sin^{2}\theta -2\sqrt{3}Sin\theta - 2Sin\theta + \sqrt{3} = 04Sin
2
θ−2
3
Sinθ−2Sinθ+
3
=0
2Sin\theta(2Sin\theta -\sqrt{3}) -1(2Sin\theta - \sqrt{3} = 02Sinθ(2Sinθ−
3
)−1(2Sinθ−
3
=0
(2Sin\theta - \sqrt{3}) (2Sin\theta - 1) = 0(2Sinθ−
3
)(2Sinθ−1)=0
If, 2Sin\theta - \sqrt{3} = 02Sinθ−
3
=0
2Sin\theta = \sqrt{3}2Sinθ=
3
Sin\theta = \dfrac{\sqrt{3} }{2}Sinθ=
2
3
Sin\theta = Sin\dfrac{\pi }{3}Sinθ=Sin
3
π
\theta = \dfrac{\pi }{3}θ=
3
π
If, 2sin\theta - 1 =02sinθ−1=0
2Sin\theta = 12Sinθ=1
Sin\theta = \dfrac{1}{2}Sinθ=
2
1
Sin\theta = Sin\dfrac{\pi }{6}Sinθ=Sin
6
π
\theta = \dfrac{\pi }{6}θ=
6
π