1) ctgx+cosx=1+ctgx*cosx
2) sin^4x+cos^4x=1/2
3) (sinx-cosx)^2=sin2x
4) sin^4x-cos^4x=7/2(sinxcosx)
5) sin^2x-sinxcosx+cosx-sinx=0
How about you try first and post your results.
2) sin^4x+cos^4x=1/2
sin^2x*sin^2x+cos^4x=1/2
(1-cos^2x)(1-cos^2x)+cos^4x=1/2
1-cos^2x-cos^2x+2cos^4x=1/2
2cos^4x-2cos^2x+1=1/2
2cos^4x-2cos^2x+1/2=0
x=pi/4+2kpi x2=7pi/4+2kpi
1) To solve the equation ctgx + cosx = 1 + ctgx * cosx, we can start by transforming it into a more manageable form:
ctgx + cosx = 1 + ctgx * cosx
ctgx - ctgx * cosx = 1 - cosx
ctgx(1 - cosx) = 1 - cosx
Now, we have two cases to consider: either ctgx = 1 - cosx or 1 - cosx = 0.
Case 1: ctgx = 1 - cosx
To find x, we can take the inverse cotangent (or arccot) of both sides:
x = arccot(1 - cosx)
Case 2: 1 - cosx = 0
Solving this equation gives us cosx = 1. From here, we can take the inverse cosine (or arccos) of both sides:
x = arccos(1)
2) To solve the equation sin^4x + cos^4x = 1/2, we can start by factoring it. First, we can notice that sin^4x and cos^4x are both perfect squares:
(sin^2x)^2 + (cos^2x)^2 = 1/2
Now, we can use the trigonometric identity sin^2x + cos^2x = 1 to simplify:
(1 - cos^2x)^2 + cos^4x = 1/2
(1 - 2cos^2x + cos^4x) + cos^4x = 1/2
1 - 2cos^2x + 2cos^4x = 1/2
Rearranging the equation:
4cos^4x - 2cos^2x + 1/2 = 0
Now, we can introduce a new variable to simplify the equation:
Let y = cos^2x
Substituting y into the equation gives us:
4y^2 - 2y + 1/2 = 0
Now, we can solve this quadratic equation for y using various methods such as factoring, completing the square, or using the quadratic formula. Once we find the possible values for y, we can substitute them back into y = cos^2x and solve for x.
3) To prove the equation (sinx - cosx)^2 = sin2x, we can expand the left side and simplify:
(sin x - cos x)^2 = (sin x - cos x)(sin x - cos x)
= (sin x - cos x)(sin x) - (sin x - cos x)(cos x)
= sin^2x - sinxcosx - sinxcosx + cos^2x
= sin^2x - 2sinxcosx + cos^2x
Next, we can use the double angle formula for sine to rewrite sin^2x and cos^2x:
= (1 - cos2x) - 2sinxcosx + (1 + cos2x)
= 1 - cos2x - 2sinxcosx + 1 + cos2x
= 2 - 2sinxcosx
Finally, we observe that 2 - 2sinxcosx is equal to sin2x:
= sin2x
Therefore, we have proven that (sinx - cosx)^2 = sin2x.
4) To solve the equation sin^4x - cos^4x = 7/2(sin xcos x), we can start by factoring the left side:
sin^4 x - cos^4 x = (sin^2 x - cos^2 x)(sin^2 x + cos^2 x)
= (sin^2 x - cos^2 x)(1)
= sin^2 x - cos^2 x
Next, we can use the Pythagorean identity sin^2 x + cos^2 x = 1 to simplify further:
= (1 - cos^2 x) - cos^2 x
= 1 - 2cos^2 x
Now, the equation becomes:
1 - 2cos^2 x = 7/2(sin xcos x)
After rearranging the equation, we have:
2 - 4cos^2 x = 7sin xcos x
Notice that the right side of the equation can be rewritten using the double angle formula for sine:
2 - 4cos^2 x = 7(2sin xcos x)
Now, we have a quadratic equation:
2 - 4cos^2 x - 14sin xcos x = 0
To solve this quadratic equation, we can use various methods such as factoring, completing the square, or using the quadratic formula. Once we find the possible values for cos x, we can substitute them back into the equation sin^2 x = 1 - cos^2 x and solve for x.
5) To solve the equation sin^2x - sinx * cosx + cosx - sinx = 0, we can try factoring it or using substitution.
Factoring:
sin^2x - sinx * cosx + cosx - sinx = 0
(sin^2x - sinx) - (cosx - sinx) = 0
sinx(sinx - 1) - (cosx - sinx) = 0
Substituting y = sinx:
y(y - 1) - (cos(x) - y) = 0
y^2 - y - cos(x) + y = 0
y^2 - cos(x) = 0
Now, we can use the identity sin^2x + cos^2x = 1 to rewrite the equation:
1 - cos(x) = 0
cos(x) = 1
Taking the inverse cosine (or arccos) of both sides, we find:
x = arccos(1)
Our final solution is x = arccos(1).