solve the equation
1. cos(θ) − sin(θ) = 1
2.2 cos(θ) tan(θ) + tan(θ) = 1 + 2 cos(θ)
3. sin(θ) cos(3θ) + cos(θ) sin(3θ) = 0
4. sin(2θ) cos(θ) − cos(2θ) sin(θ) = 1/2
5. cos(2θ) + cos(θ) = 2
6. cos(2θ) + sin2(θ) = 0
#1, look how I did this similar problem, and follow the same steps
http://www.jiskha.com/display.cgi?id=1461697741
#2
2 cos(θ) tan(θ) + tan(θ) = 1 + 2 cos(θ)
2cosθtanθ - 2cos + tanθ - 1 = 0
2cosθ(tanθ - 1) + (tanθ -1) = 0
by grouping:
(tan - 1)(2cosθ + 1) = 0
tanθ = 1 or cosθ = -1/2
θ = 45 or θ = 225 or θ = 240 or θ =300 degrees
in radians: θ = π/4, 5π/4, 4π/3, 5π/3
#3 and #4 are based on the formula
sin(A ± B) = sinAcosB ± cosAsinB
I will do #4, you do #3
sin(2θ) cos(θ) − cos(2θ) sin(θ) = 1/2
sin(2θ - θ) = 1/2
sin θ = 1/2
θ = 30 degrees or θ = 150 degrees
or θ = pi/6 or 5pi/6
#5
I will assume you mean:
cos(2θ) + sin^2 (θ) = 0
cos^2 θ - sin^2 θ + sin^2 θ = 0
cos^2 θ = 0
cosθ = 0
θ = 90 or θ = 270 degrees
θ = pi/2 or 3pi/2
To solve these equations, we can use various trigonometric identities and algebraic manipulations. Let's go through each equation step by step.
1. cos(θ) − sin(θ) = 1:
We can rewrite this equation using the Pythagorean identity sin^2(θ) + cos^2(θ) = 1:
cos(θ) − (1 − cos^2(θ)) = 1
Simplifying and rearranging terms, we get:
cos^2(θ) + cos(θ) − 2 = 0
This is a quadratic equation in terms of cos(θ). We can solve it using factoring, completing the square, or using the quadratic formula.
2. 2 cos(θ) tan(θ) + tan(θ) = 1 + 2 cos(θ):
We can start by rearranging the terms:
2 cos(θ) tan(θ) + tan(θ) - 2 cos(θ) - 1 = 0
Next, we can substitute tan(θ) with sin(θ)/cos(θ):
2 cos(θ) * (sin(θ)/cos(θ)) + (sin(θ)/cos(θ)) - 2 cos(θ) - 1 = 0
Simplifying and rearranging terms, we get:
2 sin(θ) + sin(θ) * cos(θ) - 2 cos^2(θ) - cos(θ) = 0
Using trigonometric identities, we can replace sin^2(θ) with 1 - cos^2(θ):
2 sin(θ) + sin(θ) * cos(θ) - 2(1 - cos^2(θ)) - cos(θ) = 0
Now, we have an equation in terms of cos(θ) and sin(θ). We can solve it using algebraic manipulations and techniques similar to the previous equations.
3. sin(θ) cos(3θ) + cos(θ) sin(3θ) = 0:
We can rewrite sin(3θ) and cos(3θ) using multiple angle formulas:
sin(θ) (cos(θ)cos^2(θ) - sin(θ)sin^2(θ)) + cos(θ) (cos^3(θ) - 3cos(θ)sin^2(θ)) = 0
Expanding and simplifying the equation, we get:
cos^3(θ)sin(θ) - sin^3(θ)cos(θ) - cos^2(θ)sin^2(θ) + 3cos^2(θ)sin^2(θ) = 0
We can use trigonometric identities, such as sin^2(θ) = 1 - cos^2(θ), and simplify further to solve this equation.
4. sin(2θ) cos(θ) − cos(2θ) sin(θ) = 1/2:
We can use sine and cosine double angle formula to rewrite sin(2θ) and cos(2θ):
2sin(θ)cos^2(θ) - (1 - 2sin^2(θ))sin(θ) = 1/2
Expanding and rearranging terms, we get:
2sin(θ)cos^2(θ) - sin(θ) + 2sin^3(θ) = 1/2
We can simplify and solve this equation using algebraic manipulations and trigonometric identities.
5. cos(2θ) + cos(θ) = 2:
We can use the double angle formula for cosine to rewrite cos(2θ):
2cos^2(θ) - 1 + cos(θ) = 2
Rearranging terms, we get:
2cos^2(θ) + cos(θ) - 3 = 0
This is a quadratic equation in terms of cos(θ). We can solve it using factoring, completing the square, or using the quadratic formula.
6. cos(2θ) + sin^2(θ) = 0:
We can use the Pythagorean identity sin^2(θ) + cos^2(θ) = 1 to replace sin^2(θ) with 1 - cos^2(θ):
cos(2θ) + 1 - cos^2(θ) = 0
Rearranging terms, we get:
cos^2(θ) - cos(2θ) + 1 = 0
This is a quadratic equation in terms of cos(θ). We can solve it using factoring, completing the square, or using the quadratic formula.