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Solve the equation
4sinxcosx^2 - 1 = 2cos(sin x - 1)

1 Answer

5 votes

Answer:

Explanation:

To solve this equation, we can start by using the identity cos(A-B) = cosAcosB + sinAsinB. Substituting A=sin x and B=1, we have:

2cos(sin x - 1) = cos(sin x)cos(1) + sin(sin x)sin(1)

We can then substitute this expression on the left-hand side of the original equation and simplify:

4sinxcosx^2 - 1 = cos(sin x)cos(1) + sin(sin x)sin(1)

Rearranging terms, we have:

4sinxcosx^2 - cos(sin x)cos(1) - sin(sin x)sin(1) = 1

We can now use the identity sin^2x+cos^2x=1 to simplify the left-hand side further:

4sinxcosx^2 - (sin^2x + cos^2x)cos(1) = 1

Simplifying further:

4sinxcosx^2 - (cosx)cos(1) = 1

Now we can divide both sides by cos(1):

4sinxcosx^2/cos(1) - cosx = 1/cos(1)

Rearranging terms:

cosx(4sinxcosx^2/cos(1) - 1) = 1/cos(1)

Finally, we can use the identity cos2x = 2cos^2x - 1 to simplify the left-hand side further:

(2cos^2x - 1)(4sinxcosx^2/cos(1) - 1) = 1/cos(1)

We can then divide both sides by (4sinxcosx^2/cos(1) - 1):

2cos^2x - 1 = 1/cos(1) (4sinxcosx^2/cos(1) - 1)

And finally, we can solve for cosx:

cosx = 1/2 + 1/2cos(1) (4sinxcosx^2/cos(1) - 1)

This is the solution to the equation.

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