Question

. Solve the trigonometric equation for the interval 0 ≤ 𝑥 < 2𝜋. 4 sin 𝑥 cos 𝑥 − 2√3 sin 𝑥 − 2√2 cos 𝑥 + √6 = 0 help please?

Write each value of x in each box from the smallest to the biggest

Answer 1

`2arctan(1+sqrt(2)),pi/6,-2arctan(1-sqrt(2)),11pi/6`

Explanation:

Using the so called Weierstrass Substitution

`sin(x)=2t/(1+t^2)`
`ciós(x)=(1-t^2)/(1+t^2)`
with

`tan(x/2)=t`
we get
`8t(1-t^2)/(t^2+1)^2-4sqrt(3)t/(t^2+1)-2sqrt(2)*(1-t^2)/(t^2+1)+sqrt(6)=0`
This is equivalent to

`(t-2+sqrt(3))(-t-2+sqrt(3))(-t-1+sqrt(2))(-t+1+sqrt(2)=0`
From here we get all Solutions in the given interval.