help me solve: find x in cos^2x - 2 sinx cosx - sin^2x = 0
cos ^ 2 ( x ) - sin ^ 2 ( x ) = cos ( 2x )
2 sin ( x ) cos ( x ) = sin ( 2 x )
Equation :
cos ^ 2 ( x ) - 2 sin( x ) cos( x ) -
sin ^ 2 ( x ) = 0
we can write like :
cos ( 2 x ) - sin ( 2 x ) = 0 Add sin ( 2 x ) to both sides
cos ( 2 x ) - sin ( 2 x ) + sin ( 2 x ) = 0 + sin ( 2 x )
cos ( 2 x ) = sin ( 2 x ) Divide both sides by cos ( 2 x )
cos ( 2 x ) / cos ( 2 x ) = sin ( 2 x ) / cos ( 2 x )
1 = tan ( 2 x )
tan ( 2 x ) = 1
tan ( pi / 4 ) = 1
Tangent is a periodic function with period pi , so :
2 x = n pi + pi / 4
( n is an integer ) Divide both sides by 2
x = n pi / 2 + pi / 8
To solve the equation cos^2x - 2sinx cosx - sin^2x = 0, we can rewrite it using trigonometric identities.
First, let's recall the identity: sin^2x + cos^2x = 1.
Now, let's rewrite the given equation by replacing cos^2x with 1 - sin^2x:
(1 - sin^2x) - 2sinx cosx - sin^2x = 0
Expanding and rearranging, we get:
1 - 2sin^2x - 2sinx cosx = 0
Next, we can notice that -2sin^2x - 2sinx cosx can be rewritten as -2sinx (sinx + cosx):
1 - 2sinx (sinx + cosx) = 0
Now, we can solve for x by factoring out sinx:
sinx (1 - 2(cosx + sinx)) = 0
Setting each factor equal to zero, we have two possible solutions:
sinx = 0
1 - 2(cosx + sinx) = 0
To find the values of x that satisfy sinx = 0, we know that sinx equals zero at integer multiples of π:
x = nπ, where n is an integer.
To find the values of x that satisfy 1 - 2(cosx + sinx) = 0, we solve for cosx + sinx:
cosx + sinx = 1/2
We can use the Pythagorean identity sin^2x + cos^2x = 1 to rewrite this equation:
√(2sinx)^2 + √(2cosx)^2 = 1
Simplifying, we get:
2sinx + 2cosx = 1
Rearranging, we have:
sinx + cosx = 1/2
Next, we need to use the identity for the sum of two angles:
sin(A + B) = sinAcosB + cosAsinB
Applying this identity to sinx + cosx, we have:
sin(π/4 + x) = 1/2
Now, set the angle π/4 + x equal to principal values of the sine function that yield 1/2:
π/4 + x = π/6 + 2πk or π - π/6 + 2πk, where k is an integer
Simplifying each equation, we get:
x = π/6 + 2πk - π/4 or x = π - π/6 + 2πk - π/4
Simplifying further, we have:
x = 5π/12 + 2πk or x = 7π/12 + 2πk
Therefore, the solutions for the equation cos^2x - 2sinx cosx - sin^2x = 0 are:
x = nπ, 5π/12 + 2πk, or 7π/12 + 2πk, where n and k are integers.