How to find all solutions of the equation sec(theta)cos(theta)=1
sec(theta)cos(theta)=1
(1/cosØ)(cosØ) = 1
1 = 1
Your equation is an identity, so it is true for all values of Ø, except for those values of Ø which would make cosØ = 0
e.g Ø = ±π/2, ± 3π/2 , .... would not be allowed
To find all the solutions of the equation sec(theta)cos(theta) = 1, follow these steps:
Step 1: Rewrite the equation using trigonometric identities.
sec(theta)cos(theta) = 1
Since sec(theta) is equal to 1/cos(theta), the equation can be written as:
(1/cos(theta))cos(theta) = 1
Simplifying further, we get:
1 = 1
Step 2: Notice that the equation simplifies to 1 = 1, which is a true statement.
This means that the equation is true for all values of theta.
Step 3: Write the solution.
The solution is all values of theta, or theta can take any real value.
Therefore, the equation sec(theta)cos(theta) = 1 has infinitely many solutions.
To find all solutions of the equation sec(theta)cos(theta) = 1, we need to manipulate the given equation to express it in terms of a single trigonometric function. Here's how you can do it:
1. Start by substituting sec(theta) with 1/cos(theta) in the equation:
(1/cos(theta)) * cos(theta) = 1
2. Multiply both sides of the equation by cos(theta) to eliminate the denominator:
1 = cos(theta)
3. Now, we have a simpler equation where cos(theta) equals 1.
To find all the solutions for theta, we need to consider the periodic nature of the cosine function. The cosine function has a period of 2π, which means it repeats every 2π radians or 360 degrees.
Since cos(theta) = 1 for only one specific value of theta (0 radians or 0 degrees), we need to find all the possible values of theta within one period that satisfy the equation.
To do this, we can express theta as:
θ = 2πn
where n is an integer.
Therefore, the general solution for the equation sec(theta)cos(theta) = 1 is:
θ = 2πn, where n is an integer.
This means that all values of theta that satisfy the given equation can be obtained by substituting different integer values for n. Each value of n will give a different solution for theta, and by varying n, you can find an infinite number of solutions.