Yes, by making use of the identity
cos2x = cos²x-sin²x=1-2sin²x, the given equation can be transformed to:
which is a quadratic in sin(x).
Solving the quadratic by factoring,
or sin(x)=1/2 or sin(x)=-1
Find all roots in the interval [0,2π].
The function f is continuous on the closed interval [-5,5], and f(-2) = 6, f(1) = -3, and f(4) = 6. Which of the following statements must be true? A. The equation f(x) = 0 has at least two solutions on the closed interval [-5,5].
Using inverse trigonometric functions, find a solution to the equation cos(x)=0.2 in the interval 0â‰¤xâ‰¤4Ï€. Then, use a graph to find all other solutions to this equation in this interval. Enter your answers as a
Consider the function : 3x^3 - 2x^2 - 4x + 1 Find the average slope of this function on the interval. By the Mean Value Theorem, we know there exists a "c" in the open interval (-2,3) such that f'(c) is equal to this mean slope.
Can I please get some help on these questions: 1. How many solutions does the equation,2sin^2 2 θ = sin2θ have on the interval [0, 2pi]? 4? ...what about 4cos2θ = 8cos^2 2θ? 2. True or False: sin^2 4x = 1 has 8 solutions on
We can find the solutions of sin x = 0.6 algebraically. (Round your answers to two decimal places.) (a) First we find the solutions in the interval [0, 2π). We get one such solution by taking sin−1 to get x = ________ (smaller
Please help!!!!!!!!!!! Find all solutions to the equation in the interval [0,2π). 8. cos2x=cosx 10. 2cos^2x+cosx=cos2x Solve algebraically for exact solutions in the interval [0,2π). Use your grapher only to support your
Use inverse trigonometric functions to find the solutions of the equation that are in the given interval, and approximate the solutions to four decimal places. (Enter your answers as a comma-separated list.) cos(x)(9cos(x) + 4) =
Find all solutions in the interval [0,2pi) 4sin(x)cos(x)=1 2(2sinxcosx)=1 2sin2x=1 2x=1/2 x= pi/6, and 5pi/6 Then since its 2x i divided these answers by 2 and got pi/12 and 5pi/12 However, when i checked the answer key there