solve 1-tan^2x=0 , 0≤x≤2pi
1-tan^2x=0
tan^2 x = 1
tan x = ±1
x = π/4, π - π/4, π + π/4, or 2π - π/4
x π/4, 3π/4, 5π/4, 7π/4
To solve the equation 1 - tan^2(x) = 0, we can use trigonometric identities to rewrite the equation in terms of sine and cosine. First, recall the identity:
tan^2(x) = sec^2(x) - 1
We can substitute this identity into the given equation:
1 - (sec^2(x) - 1) = 0
Next, simplify the equation by combining like terms:
1 - sec^2(x) + 1 = 0
Now, we have:
2 - sec^2(x) = 0
To solve this equation, isolate the sec^2(x) term:
sec^2(x) = 2
Now, take the square root of both sides of the equation. Remember to consider both positive and negative square roots:
sec(x) = ±√2
To find the values of x in the given interval (0 ≤ x ≤ 2π), we need to determine the values of x where sec(x) equals ±√2. We can do this by recalling the definition of secant:
sec(x) = 1/cos(x)
Therefore:
1/cos(x) = ±√2
Multiply both sides of the equation by cos(x) to isolate the variable:
1 = ±√2 * cos(x)
Now, rearrange the equation:
cos(x) = ± 1/√2
Simplify:
cos(x) = ± √2/2
Now, we need to find the values of x that satisfy this equation. By using the unit circle or the trigonometric ratios of common angles, we can determine the solutions for x within the given interval.
For the equation cos(x) = √2/2:
We know that cos(π/4) = √2/2 and cos(7π/4) = -√2/2. Therefore, the solutions for this equation within the given interval are:
x = π/4 and x = 7π/4
For the equation cos(x) = -√2/2:
We know that cos(3π/4) = -√2/2 and cos(5π/4) = √2/2. Therefore, the solutions for this equation within the given interval are:
x = 3π/4 and x = 5π/4
In conclusion, the solutions for the equation 1 - tan^2(x) = 0, within the interval 0 ≤ x ≤ 2π, are:
x = π/4, x = 3π/4, x = 5π/4, and x = 7π/4