sin^4x-cos^4x=7/2(sinxcosx)
you mean sin^4x+cos^4x=7/2(sinxcosx)
that would be much messier.
I graphed the left side as y = sin^4x+cos^4x
and the right side as y = (7/2)sinxcosx on the same grid and noticed that they intersect at
x = 15° , 75° , 195‚ and 255° , so there are actually "nice" solutions, I checked them , they work.
At the moment I don't see an easy way to actually work it out,
How about something like this
sin^2x(1-cos^2x) + cos^2x(1-sin^2x) for the left side
= sin^2x - sin^2xcos^2x + cos^2x - sin^2xcos^2x
= 1 - 2 sin^2cos^2
so 1 - 2 sin^2cos^2 = (7/2)sinxcosx
looks promising ....
2 - 4sin^2xcos^2x = 7sinxcosx
but 2sinxcosx = sin 2x
2 - (sin2x)^2 = (7/2)(2sinxcosx)
4 - 2(sin2x)^2 = 7sin2x
2sin 2x + 7sin2x - 4 = 0
(2sin2x - 1)(sin2x + 4) = 0
sin2x - 1/2 or sin2x = -4 --- > not possible
sin2x = 1/2
2x = 30 or 150
x = 15 or 75 , period of sin2x is 180, so add 180 to the answers ..... x = 195 , etc
YEAHHHH
(sin^2x + cos^2x)(sin^2x - cos^2x) = 7/2(sinxcosx)
1(sin^2x - cos^2x) = 7/2(sinxcosx)
cos^2x - sin^2x = -(7/2)(1/2)(2sinxcosx)
cos 2x = -(7/4)sin 2x
-4/7 = sin2x/cos2x
tan2x = -4/7
so 2x is in the 2nd or 4th quadrants
2x = 150.26° or 2x = 330.255
x = 75.13° or x = 165.13°
since the period of tan 2x is 90°, adding 90° to each answer gives us more answers
e.g. 165.13 + 90 = 255.13° etc
I had my calculator set to degrees, switch it to radians if you need radian answers
sin^4x+cos^4x=7/2(sinxcosx)
if it is with +..??how can u do it..can you solve it plz,cuz i have an exam
wow..thnx so much..u really are a genius
To solve the equation sin^4x - cos^4x = 7/2 * sin(x) * cos(x), we can use some trigonometric identities. Let's break it down step by step:
Step 1: Rewrite the expression using the identity sin^2x = 1 - cos^2x.
sin^4x - cos^4x = 7/2 * sin(x) * cos(x)
(sin^2x)^2 - cos^4x = 7/2 * sin(x) * cos(x)
(1 - cos^2x)^2 - cos^4x = 7/2 * sin(x) * cos(x)
Step 2: Expand and simplify the expression.
(1 - 2cos^2x + cos^4x) - cos^4x = 7/2 * sin(x) * cos(x)
1 - 2cos^2x + 2cos^4x - cos^4x = 7/2 * sin(x) * cos(x)
1 - 2cos^2x + cos^4x = 7/2 * sin(x) * cos(x)
Step 3: Rearrange the equation to have 0 on one side.
cos^4x - 2cos^2x + (1 - 7/2 * sin(x) * cos(x)) = 0
Now, solving this equation can be done by factoring or using the quadratic formula. However, let's focus on making it simpler.
Step 4: Substitute u = cos^2x.
u^2 - 2u + (1 - 7/2 * sin(x) * cos(x)) = 0
Step 5: Solve the quadratic equation.
Using the quadratic formula, the solutions for u are:
u = (2 ± √(4 - 4(1 - 7/2 * sin(x) * cos(x)))) / 2
u = 1 ± √(1 + 7sin(x)cos(x))
Step 6: Substitute u back to cos^2x.
cos^2x = 1 ± √(1 + 7sin(x)cos(x))
Step 7: Take the square root.
cos(x) = ±√(1 ± √(1 + 7sin(x)cos(x)))
Step 8: Use the Pythagorean identity sin^2x + cos^2x = 1 to substitute sin^2x.
cos(x) = ±√(1 ± √(1 + 7sin(x)√(1 - sin^2x)))
Step 9: Simplify the expression.
cos(x) = ±√(1 ± √(1 + 7sin(x)√(1 - sin^2x)))
(cos(x))^2 = 1 ± √(1 + 7sin(x)√(1 - sin^2x))
Step 10: Use the Pythagorean identity sin^2x + cos^2x = 1 to substitute sin^2x.
(cos(x))^2 = 1 ± √(1 + 7sin(x)√(cos^2x))
Step 11: Simplify the equation further.
(cos(x))^2 = 1 ± √(1 + 7cos(x))
Now, you have the equation (cos(x))^2 = 1 ± √(1 + 7cos(x)). From this point, you can solve for cos(x) using algebraic methods or numerical approximation techniques.