# How do you solve #2sin^2x-5sinx+2=0#?

Use the substitution method and then factor

Let

so,

Now you can factor

Multiply the coefficient of the first term,

Ask yourself what are the factors of

Factors of 4

Now place those factors in an order that makes it easy to factor by grouping.

Factor out

Factor out

Now you can factor out a grouping,

Now use the Zero Property

Now switch back to

Falling back to trigonometry

The side of length

All of that to say that

I have tutorials on methods of factoring found here, https://tutor.hix.ai

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To solve the equation (2\sin^2(x) - 5\sin(x) + 2 = 0), you can use a substitution method. Let (u = \sin(x)). Then the equation becomes a quadratic equation in terms of (u), which you can solve using standard quadratic equation solving techniques. Once you find the solutions for (u), you can then find the corresponding solutions for (x) by solving for (x) using the inverse sine function.

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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