I have'nt got a clue: trigonometric equations

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To solve the trigonometric equations, the first equation, sin x = (√3)/2, can be approached using the unit circle to find angles in the first and second quadrants where sine is positive, leading to solutions of x = 60° and x = 120°. For the second equation, 2 cos 2x + cos x = 0, it is recommended to use trigonometric identities to express cos 2x in terms of cos x, transforming the equation into a quadratic form. This allows for the application of standard methods to find the values of x. Understanding the properties of sine and cosine in different quadrants is crucial for identifying all possible solutions.
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I need to solve for x, but,..
honestly, i haven't got a clue for where to begin, or
even,.. oy, not sure
here are two of my problems:
sin x = (3^1/2) / 2

and

2 cos 2 x + cos x = 0

I know I'm supposed to use the unit circle,.. well, maybe
any help at all would be very much appreciated ^^
:shy:
 
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Are these equations from the same problem or are they separate? Assuming they are separate, the first one isn't hard. You already have it expressed as sin x = constant. How do you find x here?

For the second question, use a trigo identity to express cos 2x as trigo functions of x only, as opposed to 2x. Then use another trigo identity to convert the resulting expression into a quadratic equation in terms of a trigo function. You should see what to do next.
 
Wholewheat458 said:
I need to solve for x, but,..
honestly, i haven't got a clue for where to begin, or
even,.. oy, not sure
here are two of my problems:
sin x = (3^1/2) / 2

and

2 cos 2 x + cos x = 0

I know I'm supposed to use the unit circle,.. well, maybe
any help at all would be very much appreciated ^^
:shy:

The first is fairly well-known, but if you need to show it consider the unit circle. Start considering what x could be in the first quadrant. Draw a triangle OAB where O is the center of the unit circle, A is on the X axis, B is on the perimeter of the circle, OAB is a right angle and AB is \sqrt{3}/2 then x is the angle AOB. You can now find OA. Now consider if there is any way you can construct a triangle that includes the angle AOB, but where you know some of the angles. Could you for instance construct an equilateral triangle with x as an angle? (If you could then clearly x would be 60 degrees). Do the same in the second quadrant, and ask yourself whether there could be any solutions in the third and fourth quadrant (are the values of sine positive or negative in these quadrants? How about in your equation?) Don't forget to add full rotations.

For the second, try to get a quadratic equation in either cos x or sin x.
 
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