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Solve for x: 6sin^2x+2sin^2x=1 fir - 90<x<90​

Sagot :

Facundo3141592idn

Answer:

Here we have the equation:

6*sin²(x) + 2*sin²(x) = 1

and we want to find a solution in the range:

-90° < x < 90°

First, we can take the sin²(x) as a common factor to get:

6*sin²(x) + 2*sin²(x) = (6 + 2)*sin²(x) = 1

8*sin²(x) = 1

now we can divide both sides by 8

sin²(x) = 1/8

now we can apply the square root to both sides:

√(sin²(x) = √(1/8)

sin(x) = √(1/8)

Now remember the inverse sine function, Asin(x)

such that:

Asin( sin(x) ) = sin( Asin(x) ) = x

If we apply that to both sides, we get:

Asin( sin(x) )  = Asin(√(1/8))

x = Asin(√(1/8)) = 20.7°

x = 20.7°

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