By using a suitable substitution, solve the equation 4x⁴ + 7x² - 2 = 0
Hence solve the equation 4sin⁴ theta + 7sin² theta - 2 = 0 for 0° <= theta <= 360°
Let's substitute y = x² in the equation 4x⁴ + 7x² - 2 = 0.
Then we have 4(y²) + 7y - 2 = 0.
To solve this quadratic equation, we can factor it as (4y - 1)(y + 2) = 0.
Setting each factor equal to zero, we have:
4y - 1 = 0 or y + 2 = 0.
Solving for y in each equation gives us:
4y = 1 or y = -2.
So, y = 1/4 or y = -2.
Now, substitute back y = x²:
For y = 1/4, we have x² = 1/4.
Taking square root of both sides: x = ±1/2.
For y = -2, we have x² = -2.
Since the square root of a negative number is undefined, there are no real solutions in this case.
Therefore, the solutions to the equation 4x⁴ + 7x² - 2 = 0 are x = 1/2 and x = -1/2.
To solve the equation 4sin⁴ theta + 7sin² theta - 2 = 0 for 0° ≤ theta ≤ 360°, we can substitute y = sin² theta.
Then we have 4y² + 7y - 2 = 0.
To solve this quadratic equation, we can factor it as (4y - 1)(y + 2) = 0.
Setting each factor equal to zero, we have:
4y - 1 = 0 or y + 2 = 0.
Solving for y in each equation gives us:
4y = 1 or y = -2.
So, y = 1/4 or y = -2.
Now, substitute back y = sin² theta:
For y = 1/4, we have sin² theta = 1/4.
Taking square root of both sides: sin theta = ±1/2.
For y = -2, there are no real solutions since sin² theta cannot be negative.
Therefore, the solutions to the equation 4sin⁴ theta + 7sin² theta - 2 = 0 for 0° ≤ theta ≤ 360° are:
theta = 30°, 150°, 210°, and 330°.
To solve the equation 4x⁴ + 7x² - 2 = 0, we can use a substitution. Let's substitute x² with a new variable, let's say u. So, u = x².
Now, let's rewrite the equation using the substitution:
4u² + 7u - 2 = 0.
Now, we can solve this quadratic equation for u:
4u² + 7u - 2 = 0.
To factorize this equation, we look for two numbers whose product is -8 and whose sum is 7. The numbers are 8 and -1:
(2u - 1)(2u + 2) = 0.
Now, setting each factor equal to zero and solving for u:
2u - 1 = 0 or 2u + 2 = 0.
For the first equation, solving for u:
2u - 1 = 0.
2u = 1,
u = 1/2.
For the second equation, solving for u:
2u + 2 = 0.
2u = -2,
u = -1.
Now, substitute back u = x²:
For u = 1/2,
x² = 1/2,
x = ±√(1/2)
x = ±(1/√2)
x = ±(√2/2)
For u = -1,
x² = -1,
Since we can't take the square root of a negative number, there are no real solutions for this case.
Therefore, the solutions to the equation 4x⁴ + 7x² - 2 = 0 are:
x = √2/2 or x = -√2/2.
Now, let's solve the equation 4sin⁴ theta + 7sin² theta - 2 = 0 for 0° <= theta <= 360° using the values we found for x:
For x = √2/2,
sin² theta = (√2/2)²,
sin² theta = 1/2,
sin theta = ±√(1/2),
theta = 45° or 135°.
For x = -√2/2,
sin² theta = (-√2/2)²,
sin² theta = 1/2,
sin theta = ±√(1/2),
theta = 45° or 135°.
Therefore, the solutions to the equation 4sin⁴ theta + 7sin² theta - 2 = 0 for 0° <= theta <= 360° are:
theta = 45°, 135°.