Discussion Topic

# Solving Trigonometric Equations

Summary:

To solve trigonometric equations, first isolate the trigonometric function. Next, determine the general solution using inverse trigonometric functions. Finally, consider the specific interval given in the problem and use the unit circle to find all possible solutions within that interval.

Solve the equation 5cos^2x-6tanx*cosx=cos^2x-5sin^2x-1.

We have to solve 5(cos x)^2 - 6 tan x*cos x = (cos x)^2- 5*( sin x)^2 - 1

5(cos x)^2 - 6 tan x*cos x = (cos x)^2- 5*( sin x)^2 - 1

=> 5(cos x)^2 + 5*( sin x)^2 - 6 tan x*cos x = (cos x)^2 - 1

we know (cos x)^2 + ( sin x)^2 = 1

=> 5 - 6 tan x*cos x = (cos x)^2 - 1

=> 6 - 6 (sin x/ cos x)* cos x =  (cos x)^2

=> 6 - 6 sin x = 1 - (sin x)^2

let y = sin x

=> 6 - 6y = 1 - y^2

=> y^2 - 6y+ 5 = 0

=> y^2 - 5y - y + 5 = 0

=> y( y - 5) - 1(y - 5) = 0

=> (y - 1)(y - 5) = 0

y = 1 or 5

As y = sin x, we can ignore x = 5.

So sin x = 1

x = arc sin (1)

x = pi/2 + 2*n*pi

Therefore x = pi/2 + 2*n*pi

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Solve the equation 5sinx=4cosx

To solve the equation 5sinx = 4cosx, first, bring all terms to one side of the equations, so that the other side becomes 0:

5sinx - 4cosx = 0.

Then, factor out 4cosx from the right hand side:

4cosx((5sinx)/(4cosx) - 1) = 0.

The product equals zero when either of the factors is zero, so

cos(x) = 0 or (5sinx)/(4cosx) - 1 = 0.

If cos(x) = 0, then sin(x) also has to be zero in order for the original equation to be true. However, it is impossible for cos(x) and sin(x) to be zero simultaneously (for the same values of x). Therefore, cos(x) = o does not yield valid solutions.

If (5sinx)/(4cosx) - 1 = 0, then

5/4 tan(x) = 1, or

tan(x) = 4/5

This equation has solutions of the form

x = arctan(4/5) +pi*k, where k is an integer.

Dividing the original equation 5sinx = 4cosx by cosx will also result in the correct answer in this case. However, dividing equation by a variable or a function that can have the value of zero should be done with caution, as it could result in missing some of the possible solutions. In this case, cosx cannot be zero, as shown in the solution above, so dividing by cosx does not present a problem.

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Solve the equation 5sinx=4cosx

The previous educator's post is a perfectly reasonable, conventional way of solving the problem of 5 sin x = 4 cos x. Namely, to recognize that the tangent function (tanx) is equivalent to (sinx)/(cosx), and then simply defining the solution as arctan(4/5). I would add that arctan might appear on a graphing utility as tan^-1(x), and is not to be confused with cot(x).

Next, I would add that this problem can be solved with a graphing utility by moving both functions to one side of the equation and graphing the function, as follows:

5 sin x - 4 cos x = 0. This results in intercepts of .675, 3.816, 6.958, etc.

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Solve the equation 5sinx=4cosx

We have to solve 5 sin x = 4 cos x

5 sin x = 4 cos x

=> sin x / cos x = 4/5

=> tan x = (4/5)

=> x = arc tan (4/5)

Therefore x = arc tan (4/5) + n*pi