
Giải pt; sin^4 (x/3) + cos^4 (x/3)=5/8


Ответы на вопрос

sin^4(x/3)+cos^4(x/3)= 1-2sin^2(x/3)*cos^2(x/3)= 1 - 0,5sin^2(2x/3).
Вернемся уравнению:
1-0,5sin^2(2x/3)=5/8
0,5sin^2(2x/3)=3/8
sin^2(2x/3)=6/8=3/4
Тогда sin(2x/3)=sqrt(3)/2 или sin(2x/3)= -sqrt(3)/2
sin(2x/3)=sqrt(3)/2 => 2x/3=(-1)^k * pi/3 + pi * k;
x=(-1)^k * pi/2 + 3pi/2 * k, где k - целое число
sin(2x/3)= -sqrt(3)/2 => 2x/3=(-1)^n * (-pi/3)+pi * n;
x=(-1)^(n+1) * pi/2 + 3pi/2 * n, где n - целое число



To solve the equation sin^4(x/3) + cos^4(x/3) = 5/8, we can use some trigonometric identities and algebraic manipulation. Let's go through the steps:
Step 1: Recall the trigonometric identity for cosine and sine squared: sin^2(x) + cos^2(x) = 1
Step 2: Rewrite the given equation using the above identity: (sin^2(x/3))^2 + (cos^2(x/3))^2 = 5/8
Step 3: Now, we'll use another trigonometric identity: sin^2(x) = 1 - cos^2(x)
Step 4: Substitute sin^2(x/3) in terms of cos^2(x/3) using the above identity: (1 - cos^2(x/3))^2 + (cos^2(x/3))^2 = 5/8
Step 5: Expand and simplify the equation: 1 - 2*cos^2(x/3) + cos^4(x/3) + cos^2(x/3) = 5/8
Step 6: Rearrange the terms and combine like terms: cos^4(x/3) - cos^2(x/3) + 3/8 = 0
Step 7: Let's make a substitution to simplify the equation further: Let y = cos^2(x/3)
Now the equation becomes: y^2 - y + 3/8 = 0
Step 8: Solve the quadratic equation for y. To do this, we can use the quadratic formula: y = (-b ± √(b^2 - 4ac)) / 2a
where a = 1, b = -1, and c = 3/8
y = (1 ± √(1 - 4*(1)*(3/8))) / 2 y = (1 ± √(1 - 3/2)) / 2 y = (1 ± √(2/2 - 3/2)) / 2 y = (1 ± √(-1/2)) / 2
Step 9: Since the value of y cannot be negative (because cos^2(x/3) cannot be negative), we discard the negative solution: y = (1 + √(-1/2)) / 2
Step 10: Now, we encounter a problem because the term √(-1/2) results in an imaginary number. This means that the given equation has no real solutions. It is not possible to find real values of x that satisfy sin^4(x/3) + cos^4(x/3) = 5/8.
Hence, the solution to the equation sin^4(x/3) + cos^4(x/3) = 5/8 is "No real solutions."


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