Question

Maths question

Answer 1

Solve the following eqn for −π≤ x≤π , in terms of π

Answer 2

\(4sinx-7=2cosecx\)

Answer 3

My woRKING:
\(4sin^2-7=2cosecx\)
\(4sin^2x-7sinx-2=0\)
\((sinx-2)(4sinx+1)=0\)
\(sinx=2(no~solution)\),\(sinx=\frac{-1}{4}\)

Answer 4

\(\sin x=\frac{-1}{4} \\ x = \sin^{-1} \left( \frac{-1}{4}\right) + 2\pi n,~ \pi - \sin^{-1} \left( \frac{-1}{4}\right) + 2\pi n\)
\(x=1.08\pi+2\pi~n=1.08\pi+2\pi(-1)\approx-0.92\pi\)
\(x=1.92\pi+2\pi~n=1.92\pi+2\pi(-1)\approx-0.08\pi\)

Answer 5

Is my working correct or...?

Answer 6

@mathmale