Expert Answer to "Solving 4sin4x−4sin2x+1=0,x∈[0,2π⟩ I am having problem solving this..."

Jamya Elliott

Answered question

2022-01-23

Solving

4 \sin^{4} x - 4 \sin^{2} x + 1 = 0, x \in [0, 2 π ⟩

I am having problem solving this trig equation. Every time I try to simplify or substitute sinx with u, I get

x = \emptyset

.
How do I go about solving this equation?

4 \sin^{4} x - 4 \sin^{2} x + 1 = 0, x \in [0, 2 π ⟩

tipoule137p

Beginner2022-01-24Added 10 answers

Substitute

\sin^{2} (x) = t

and solve the quadratic.

Wilson Mitchell

Beginner2022-01-25Added 8 answers

The left-hand side is

{(2 \sin^{2} x - 1)}^{2}

, so you need to solve

\sin^{2} x = \frac{12}{}

(i.e.

\cos 2 x = 1 - 2 \sin^{2} x = 0

) for

2 x \in [0, 4 π)

. The solution is therefore

x \in {\frac{π}{4}, \frac{3 π}{4}, \frac{5 π}{4}, \frac{7 π}{4}}

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