(c) sin(2x) − 4 sin(x) = 0 simplify
2sinxcosx - 4sinx = 0 factor
2sinx [ cosx - 2] = 0 the second factor has no solution
So
2sinx = 0 divide both sides by 2
sinx = 0 and this occurs at [0 + n*180] ° where n is an integer
(d) 4cos2(θ)−cos(θ)−2=0 simplify
4[2cos^2 (θ) - 1] - cos (θ) - 2 = 0
8cos^2 (θ) - cos (θ) - 6 = 0
Let x = cos(θ)
8x^2 - x - 6 = 0
Using the quadratic formula, the solutions to this are x = [1 - sqrt(193)] /16 and x = [1 + sqrt(193)]/ 16
So
cos (θ) = [1 - sqrt(193)] /16 or cos (θ ) = [1 +sqrt(193)] /16
Taking the cosine inverse to find both angles, we have
(θ) = about 143.685° and (θ) = about 21.444°
There are also two more angles on (0, 360) at about 216.3° and about 338.6°
Here's the graph of this one......it has an odd periodicity.......https://www.desmos.com/calculator/pwgqsyrurd
(c) sin(2x) − 4 sin(x) = 0 simplify
2sinxcosx - 4sinx = 0 factor
2sinx [ cosx - 2] = 0 the second factor has no solution
So
2sinx = 0 divide both sides by 2
sinx = 0 and this occurs at [0 + n*180] ° where n is an integer
(d) 4cos2(θ)−cos(θ)−2=0 simplify
4[2cos^2 (θ) - 1] - cos (θ) - 2 = 0
8cos^2 (θ) - cos (θ) - 6 = 0
Let x = cos(θ)
8x^2 - x - 6 = 0
Using the quadratic formula, the solutions to this are x = [1 - sqrt(193)] /16 and x = [1 + sqrt(193)]/ 16
So
cos (θ) = [1 - sqrt(193)] /16 or cos (θ ) = [1 +sqrt(193)] /16
Taking the cosine inverse to find both angles, we have
(θ) = about 143.685° and (θ) = about 21.444°
There are also two more angles on (0, 360) at about 216.3° and about 338.6°
Here's the graph of this one......it has an odd periodicity.......https://www.desmos.com/calculator/pwgqsyrurd