I am almost finished iwth my questionings:)
Solve the given triginometric equation analytically
sin4x - cos2x = 0
Priya
let's use the identity
sin2A = 2sinAcosA on the first term only
2sin2xcos2x - cos2x = 0 , now a common factor
cos2x(2sin2x - 1) = 0
cos2x = 0 or sin2x = 1/2
so 2x = 90, 270, 30, 150º
then x = 45,135,15,75º
but the period of sin2x and cos2x is 180º
so other answers are 225,315,195º
to change these quickly to radians, take the smallest degree angle which is 15º
15º = pi/12 radians
all other angles are multiples of 15
eg. 315 = 21*15
so 315º = 21pi/12 etc.
Given angle 1 = (x + 7)degree, angle 2 = *(2x -3)degree, angle ABC = (x 2nd power) degree, angle D = (5x - 4)degree. Show that angle ABC congruent to angle D
To solve the trigonometric equation sin(4x) - cos(2x) = 0 analytically, we need to manipulate the equation to isolate the variable x.
Let's start by rewriting the equation using trigonometric identities. We know that sin(2x) = 2sin(x)cos(x), so we can substitute that into the equation:
sin(4x) - cos(2x) = 0
2sin(2x)cos(2x) - cos(2x) = 0
Now, let's factor out the common factor of cos(2x):
cos(2x)(2sin(2x) - 1) = 0
To solve this equation, we analyze the two factors separately:
1. cos(2x) = 0
To find the solutions for this factor, we can set 2x equal to the cosine inverse of 0, which is π/2 or 3π/2. Therefore, 2x = π/2 + kπ or 2x = 3π/2 + kπ, where k is an integer.
Solving for x, we divide both sides by 2:
x = π/4 + kπ/2 or x = 3π/4 + kπ/2, where k is an integer.
2. 2sin(2x) - 1 = 0
Add 1 to both sides and divide by 2:
2sin(2x) = 1
sin(2x) = 1/2
To find the solutions for this factor, we can use the inverse sine function. The solution will be in the range [-π/2, π/2]. So we have:
2x = π/6 or 2x = 5π/6 + 2πn, where n is an integer.
Solving for x, we divide both sides by 2:
x = π/12 or x = 5π/12 + πn, where n is an integer.
Therefore, the solution for the trigonometric equation sin(4x) - cos(2x) = 0 is:
x = π/4 + kπ/2, x = 3π/4 + kπ/2, x = π/12 + πn, or x = 5π/12 + πn, where k and n are integers.