Question

2*sin(2x+pi/6)+cosx = sqrt(3)*sin(2x)-1

Answer 1

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Answer 2

To solve the equation 2sin(2x+π/6) + cos(x) = √3sin(2x) - 1, we'll simplify it step by step.

Let's start by expanding the trigonometric functions:

2sin(2x)cos(π/6) + cos(x) = √3sin(2x) - 1

Using the identity sin(2θ) = 2*sin(θ)*cos(θ), we can simplify the equation further:

sin(π/3)sin(2x) + cos(x) = √3*sin(2x) - 1

Since sin(π/3) = √3/2, we can substitute the value:

(√3/2)sin(2x) + cos(x) = √3sin(2x) - 1

Next, let's isolate the sin(2x) terms on one side of the equation by subtracting (√3/2)*sin(2x) from both sides:

cos(x) - (√3/2)*sin(2x) = -1

Now, let's rewrite sin(2x) in terms of sin(x) and cos(x) using the double-angle identity:

cos(x) - (√3/2)(2sin(x)*cos(x)) = -1

Expanding the expression:

cos(x) - √3*sin(x)*cos(x) = -1

Next, we can factor out cos(x) from the left side:

cos(x)(1 - √3sin(x)) = -1

Now, divide both sides by (1 - √3*sin(x)):

cos(x) = -1 / (1 - √3*sin(x))

We have simplified the equation, but finding an explicit solution might not be straightforward due to the transcendental nature of trigonometric functions. You could try numerical methods or approximations to find an approximate solution for x.