# How do you use the double angle or half angle formulas to solve #2 sinx cosx=cos 2x#?

Solve 2sin x.cos x = cos 2x

Ans:

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To solve the equation ( 2 \sin(x) \cos(x) = \cos(2x) ) using double angle or half-angle formulas, we can start by using the double angle identity for cosine:

[ \cos(2x) = \cos^2(x) - \sin^2(x) ]

Now, we substitute this expression into the given equation:

[ 2 \sin(x) \cos(x) = \cos^2(x) - \sin^2(x) ]

Next, we can use the identity ( \sin^2(x) = 1 - \cos^2(x) ) to replace ( \sin^2(x) ) in the equation:

[ 2 \sin(x) \cos(x) = \cos^2(x) - (1 - \cos^2(x)) ]

[ 2 \sin(x) \cos(x) = \cos^2(x) - 1 + \cos^2(x) ]

Now, we simplify the equation:

[ 2 \sin(x) \cos(x) = 2 \cos^2(x) - 1 ]

[ 2 \sin(x) \cos(x) = 2 \cos^2(x) - 1 ]

[ 2 \sin(x) \cos(x) - 2 \cos^2(x) + 1 = 0 ]

Now, let's factor out ( \cos(x) ) from the first two terms:

[ 2 \cos(x)(\sin(x) - \cos(x)) + 1 = 0 ]

Now, we have a quadratic equation in terms of ( \cos(x) ). We can solve it using the quadratic formula or factoring techniques.

After finding the values of ( \cos(x) ), we can then use the inverse cosine function to find the values of ( x ) within the desired range (typically ( 0 \leq x \leq 2\pi ) or ( 0^\circ \leq x \leq 360^\circ )).

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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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