Please Help
Solve each equation for 0<x<2 pi
2cos^ 2 x +sins-1=0
Please proof read.
Solve equation for 0< Equality x < 2pi
Trig equations
I'm not sure if this is right but this is what I have so far
2cos^2 +sinx -1 =0
2cos ^2 x+sinx =1 (I added 1 to both sides)
Cos^2 x +sinx
you have to keep writing an equation; you just dropped off the right side. In any case, remember from your Algebra I that you usually need to set stuff to zero to solve.
Now, using the fact that sin^2x+cos^2x=1,
2cos^2x+sinx-1 = 0
2-2sin^2x + sinx-1 = 0
2sin^2x-sinx-1 = 0
(2sinx+1)(sinx-1) = 0
sinx = -1/2 or 1
Now list the angles in all the quadrants that fit.
To solve the equation 2cos^2(x) + sin^(-1)(x) = 0 for 0 < x < 2pi, we need to simplify the equation and then find the values of x that make it equal to zero.
Let's start by simplifying the equation step by step:
1. Replace the sin^(-1)(x) term with its equivalent, which is arcsin(x):
2cos^2(x) + arcsin(x) = 0
2. Since we know that 0 < x < 2pi, we need to find the values of x within this range that satisfy the equation.
3. Rearrange the equation to isolate the cosine term:
2cos^2(x) = -arcsin(x)
4. Divide both sides of the equation by 2 to get cos^2(x) alone:
cos^2(x) = -arcsin(x)/2
5. Take the square root of both sides to solve for cos(x):
cos(x) = sqrt(-arcsin(x)/2)
6. Notice that there is a square root of a negative number, which is not defined in the real number system. Therefore, there are no real solutions to this equation.
In conclusion, there are no values of x within the range 0 < x < 2pi that satisfy the equation 2cos^2(x) + sin^(-1)(x) = 0.