Help with Trigonometry equation

In summary, the algebraic method to get from X = (pi)/4 + 2(pi)K into the form X = (pi)/4 + (pi)/2K involves finding all the solutions to the general equation 6 sin^2 X-3=0, where X is in the range [0, 2pi), and modifying the solution to include all possible values of X, which is represented by adding (pi)/2K. This is because sin is a periodic function and adding multiples of (pi)/2 to the solution will still give valid solutions. Therefore, the final equation is x=(pi/4)+(pi/2)*K, where K is any integer.
  • #1
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What is the algebraic method to get from X = (pi)/4 + 2(pi)K into this form => X = (pi)/4 + (pi)/2K ? These are the general equations for all solutions to this original equation: 6 sin^2 X-3=0 , I can understand how to get the general equation graphically can someone show me how to get from the first general equation to the second general equation which is the correct version?
 
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  • #2
when you solve 6*sin^2(x)-3=0, you get sin(x)=+/- 1/sqrt(2). So, the first answer you gave is x=(pi/4)+2*pi*K. But this isn't all the answers. There are other such x values in [0,2pi) that satisfy sin(x)=+/- 1/sqrt(2) and they are (3pi)/4, (5pi)/4 and (7pi)/4. So you need to modify your solution to start with pi/4 and add these other values and all their multiples, which basically comes down to your second solution x=(pi/4)+(pi/2)*K since (3pi)/4=(pi/4)+(pi/2)*1, (5pi)/4=(pi/4)+(pi/2)*2 and so on.
 
  • #3
i guess the solution needs to be modified since they are all multiples of pi/4. There might be a better way to understand this , I solved
3*pi/4 = pi/4 + 2*pi*k for k and got k=1/4. Then plugged this into the general equation for all solutions and got, pi/4+2*pi*1/4 which becomes
pi/4+pi/2 and just stuck a k on the end to get the general solution of
pi/4+pi/2*k. Is this correct reasoning because I don't understand why the k belongs on the end, because I already solved for K, it seems like I am using intuition rather than a logical algebraic method for why it goes back on the end.
 
  • #4
in your last post, you are showing algebraically what K needs to be in order to get (3pi)/4. But the K must stay in the equation because sin is a periodic function. Like I said, pi/4, (3pi)/4, (5pi)/4 and (7pi)/4 all work as solutions. But if you add 2pi to any of these they always work. So the equation that accounts for all of this is x=(pi/4)+(pi/2)*K, where K is any integer, since to get from one solution to the next you just have to add some multiple of (pi/2).

so to sum it all up: the equation x=(pi/4)+2*pi*K is a correct equation but if K can only be an integer, it doesn't get all of the solutions. Like you showed, in order to get (3pi)/4 you need K=1/4. So rather than keeping that equation where this K can be fractions, they changed the equation to x=(pi/4)+(pi/2)*K, where K is any integer, which gives all the solutions in the most simple manner. There is not a way to algebraically manipulate the first answer to get the second one. It is just the reasoning of what the solution actually is and what the best way to present it would be.
 

FAQ: Help with Trigonometry equation

What is Trigonometry?

Trigonometry is a branch of mathematics that deals with the relationships between the sides and angles of triangles. It is used to solve problems involving right triangles and can also be applied to non-right triangles.

How do I solve a trigonometry equation?

To solve a trigonometry equation, you will need to use the trigonometric functions of sine, cosine, and tangent. You will also need to know the values of the sides and angles of the triangle. You can then use the appropriate trigonometric function to find the missing side or angle.

What are the common trigonometric functions?

The most common trigonometric functions are sine, cosine, and tangent. These functions are abbreviated as sin, cos, and tan respectively. Other common functions include cosecant (csc), secant (sec), and cotangent (cot).

How do I remember the trigonometric identities?

One way to remember the trigonometric identities is by using the acronym "SOH-CAH-TOA", which stands for sine = opposite/hypotenuse, cosine = adjacent/hypotenuse, and tangent = opposite/adjacent. You can also create flashcards or practice problems to help memorize the identities.

How can I use trigonometry in real life?

Trigonometry is used in a variety of real-life applications, such as architecture, engineering, navigation, and astronomy. It can be used to calculate the height of a building, the distance between two points, or the trajectory of a projectile. It is also used in fields such as music and art to create patterns and designs.

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