How do I solve these?
1) 2sinxcosx-cosx=0
2) cos^2(x)-0.5cosx=0
3) 6sin^2(x)-5sinx+1=0
4) tan^2(x)+tanx-12=0
Factor them and set each of the factors equal to zero. Examples:
(1) cos x(2 sinx -1) = 0
cos x = 0 or sin x = 0.5
x = pi/2, 3 pi/2; pi/6, 5 pi/6
(4) (tanx -3)(tanx +4) = 0
x = tan^-1 (1/3) or tan^-1(-1/4)
My answers above are in radians. You can easily convert them to degrees, if that is what they want.
pi/2 radians it 90 degrees, etc
1) To solve the equation 2sinxcosx - cosx = 0:
Step 1: Factor out the common factor cosx:
cosx(2sinx - 1) = 0
Step 2: Set each factor equal to 0 and solve for x:
cosx = 0, giving us x = π/2 + nπ or x = 3π/2 + nπ, where n is an integer.
2sinx - 1 = 0, giving us sinx = 1/2. Solving for x, we have x = π/6 + nπ or x = 5π/6 + nπ, where n is an integer.
2) To solve the equation cos^2(x) - 0.5cosx = 0:
Step 1: Factor out the common factor cosx:
cosx(cosx - 0.5) = 0
Step 2: Set each factor equal to 0 and solve for x:
cosx = 0 gives us x = π/2 + nπ or x = 3π/2 + nπ, where n is an integer.
cosx - 0.5 = 0 gives us cosx = 0.5. Solving for x, we have x = π/3 + 2nπ or x = 5π/3 + 2nπ, where n is an integer.
3) To solve the equation 6sin^2(x) - 5sinx + 1 = 0:
This equation can be factorized. We have:
(2sinx - 1)(3sinx - 1) = 0
Setting each factor equal to 0 and solving for x, we get:
2sinx - 1 = 0 gives us sinx = 1/2. Solving for x, we have x = π/6 + nπ or x = 5π/6 + nπ, where n is an integer.
3sinx - 1 = 0 gives us sinx = 1/3. Solving for x, we have x = arcsin(1/3) + 2nπ or x = π - arcsin(1/3) + 2nπ, where n is an integer.
4) To solve the equation tan^2(x) + tanx - 12 = 0:
Step 1: Factorize the quadratic equation:
(tanx - 3)(tanx + 4) = 0
Step 2: Set each factor equal to 0 and solve for x:
tanx - 3 = 0 gives us tanx = 3. Solving for x, we have x = arctan(3) + nπ, where n is an integer.
tanx + 4 = 0 gives us tanx = -4. Solving for x, we have x = arctan(-4) + nπ, where n is an integer.
To solve these equations, we will use algebraic techniques and trigonometric identities. Let's go through each equation step by step:
1) 2sin(x)cos(x) - cos(x) = 0:
First, we can factor out the common factor of cos(x):
cos(x) (2sin(x) - 1) = 0
Now we have two possibilities:
a) cos(x) = 0: This occurs when x is equal to π/2 or 3π/2 (using the unit circle or the graph of the cosine function).
b) 2sin(x) - 1 = 0: Solving this equation for sin(x), we get sin(x) = 1/2, which occurs when x is equal to π/6 or 5π/6.
So, the solutions to the equation 2sin(x)cos(x) - cos(x) = 0 are x = π/2, 3π/2, π/6, and 5π/6.
2) cos^2(x) - 0.5cos(x) = 0:
We can rearrange this equation as follows:
cos(x) (cos(x) - 0.5) = 0
Again, we have two possibilities:
a) cos(x) = 0: This happens when x is equal to π/2 or 3π/2.
b) cos(x) - 0.5 = 0: Solving this equation for cos(x), we get cos(x) = 0.5, which occurs when x is equal to π/3 or 5π/3.
The solutions to the equation cos^2(x) - 0.5cos(x) = 0 are x = π/2, 3π/2, π/3, and 5π/3.
3) 6sin^2(x) - 5sin(x) + 1 = 0:
This quadratic equation can be factored as follows:
(2sin(x) - 1)(3sin(x) - 1) = 0
Solving the first factor, we get:
2sin(x) - 1 = 0
sin(x) = 1/2
This occurs when x is equal to π/6 or 5π/6.
Solving the second factor, we get:
3sin(x) - 1 = 0
sin(x) = 1/3
This occurs when x is equal to arcsin(1/3) or π - arcsin(1/3).
So, the solutions to the equation 6sin^2(x) - 5sin(x) + 1 = 0 are x = π/6, 5π/6, arcsin(1/3), and π - arcsin(1/3).
4) tan^2(x) + tan(x) - 12 = 0:
This equation is a quadratic equation in terms of tan(x). Let's solve it by factoring:
(tan(x) + 4)(tan(x) - 3) = 0
Solving the first factor, we get:
tan(x) + 4 = 0
tan(x) = -4
This occurs when x is equal to arctan(-4) or π + arctan(-4).
Solving the second factor, we get:
tan(x) - 3 = 0
tan(x) = 3
This occurs when x is equal to arctan(3).
So, the solutions to the equation tan^2(x) + tan(x) - 12 = 0 are x = arctan(-4), π + arctan(-4), and arctan(3).
Remember to consider periodicity and the restricted domain of trigonometric functions when finding all solutions.