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°.
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°
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