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Find all real numbers in the interval [0,2pi) that satisfy the equation. 4sin^(2)gamma+1sin gamma-5=0?

Sagot :

Answer: (Ill name gamma x to make equation more understandable).

[tex]4sin^{2}x + sin x - 5 = 0
\\ \\ \\
4sin^{2}x + sin x = 5
\\ \\ \\
[/tex]

We see that [tex]sin(\frac{\pi}{2})[/tex] makes in both cases one, so:

[tex]4sin^{2}\frac{pi}{2} + sin \frac{pi}{2} =
\\ \\ \\
4 \cdot 1 + 1 \cdot 1 = 4 + 1 = 5[/tex]

So answer could be [tex]\boxed{\frac{pi}{2}}[/tex], but there can be added any constant and answer will still be 1. So main solution is: [tex]\boxed{\frac{pi}{2} + 2k \pi}[/tex], where k is any integer.

Hope this helps!:)