The only real solution is theta = 0
For theta < 0 square root of 3 theta is not defined.
For theta > 0, sin theta increases slower than 3*theta and so the sum is always negative.
The only real solution is theta = 0
For theta < 0 square root of 3 theta is not defined.
For theta > 0, sin theta increases slower than 3*theta and so the sum is always negative.
The only real solution is theta = 0
For theta < 0 square root of 3 theta is not defined.
For theta > 0, sin theta increases slower than 3*theta and so the sum is always negative.
The only real solution is theta = 0
For theta < 0 square root of 3 theta is not defined.
For theta > 0, sin theta increases slower than 3*theta and so the sum is always negative.
87
Take the square root of both sides. x=the square root of 10. It is an irrational number. Approximately 3.15
To solve the equation (\sin^2 \theta = 0.75), first take the square root of both sides to get (\sin \theta = \pm \sqrt{0.75} = \pm \frac{\sqrt{3}}{2}). Then, find the angles (\theta) for which (\sin \theta = \frac{\sqrt{3}}{2}) and (\sin \theta = -\frac{\sqrt{3}}{2}). The solutions are (\theta = \frac{\pi}{3} + 2k\pi) and (\theta = \frac{2\pi}{3} + 2k\pi) for the positive case, and (\theta = \frac{7\pi}{6} + 2k\pi) and (\theta = \frac{4\pi}{3} + 2k\pi) for the negative case, where (k) is any integer.
This question cannot be answered. You will have to give me the number to the square root. * * * * * a = ±sqrt(c^2 - b^2)
Square root both sides and then x = 4
If tan theta equals 2, then the sides of the triangle could be -2, -1, and square root of 5 (I used the Pythagorean Theorem to get this). From this, sec theta is negative square root of 5. It is negative because theta is in the third quadrant, where cosine, secant, sine, and cosecant are all negative.
If x equals the square root of ...., then you already have solved for x
Negative 1.047197551 etc, etc.
87
68
It depends on which variable you wish to solve for.
Take the square root of both sides. x=the square root of 10. It is an irrational number. Approximately 3.15
To solve the equation (\sin^2 \theta = 0.75), first take the square root of both sides to get (\sin \theta = \pm \sqrt{0.75} = \pm \frac{\sqrt{3}}{2}). Then, find the angles (\theta) for which (\sin \theta = \frac{\sqrt{3}}{2}) and (\sin \theta = -\frac{\sqrt{3}}{2}). The solutions are (\theta = \frac{\pi}{3} + 2k\pi) and (\theta = \frac{2\pi}{3} + 2k\pi) for the positive case, and (\theta = \frac{7\pi}{6} + 2k\pi) and (\theta = \frac{4\pi}{3} + 2k\pi) for the negative case, where (k) is any integer.
This question cannot be answered. You will have to give me the number to the square root. * * * * * a = ±sqrt(c^2 - b^2)
Square root both sides and then x = 4
x2=-81take the square root of both sidessince 81 is negative you need to take 81 times i when i=-1 then solve as the square root of 81. your answer is x=9i
sine[theta]=opposite/hypotenuse=square root of (1-[cos[theta]]^2)
Yes and x = 2+square root of 6 or x = 2-square root of 6
No, pi is not used to solve a square root problem.
It doesn't factor neatly. If you solve it, you get the square root of 12/7 which is either 2 times the square root of 3/7 or -2 times the square root of 3/7
Square root of 6.25 equals ± 2.5
Take the square root of both sides of the equation
2x2 = 228 divide both sides by 2 x2 = 114 x = ±√114 by the square root property
What square root property is essential to solve any radical equation involving square root?
sqrtx=20 to get rid of the square root you square both sides so sqrt(x)^2=x and (20)^2=400 x=400