Solving Trigonometric Equations

We will solve the given equation by first isolating sin θ on the left-hand side.

Next, we will apply the inverse of the sine function on both sides of the equation.

We found that one solution to the equation is θ =- 30^(∘). However, we want the solutions to be in the interval from 0^(∘) to 360^(∘). Since the solution we found is negative, the angle is measured clockwise. To find the solution in the desired interval, we subtract 30 from 360.

The first solution in the desired interval is θ=330^(∘). To find other possible solutions, recall that the sine of an angle is the y-coordinate of the point of intersection P between the unit circle and the terminal side of the angle. P(x,y) = (cosθ, sinθ) Therefore, we will find another angle whose terminal side intersects the unit circle at a point that has a y-coordinate of - 12.

In the diagram, we can see that another solution to the equation is θ=210^(∘). In fact, we see that, in the desired interval, the only solutions to the equation are θ = 210^(∘) and θ=330^(∘).

Let's start by isolating cos θ on the left-hand side.

Next, we will apply the inverse of the cosine function on both sides of the equation.

We found that one solution to the equation is θ =60^(∘), which is in the interval from 0^(∘) to 360^(∘). Since the solution we found is positive, the angle is measured counterclockwise.

To find other possible solutions, recall that the cosine of an angle is the x-coordinate of the point of intersection P between the unit circle and the terminal side of the angle. P(x,y) = ( cosθ,sinθ) Therefore, we will find another angle whose terminal side intersects the unit circle at a point that has an x-coordinate of 12.

In the diagram, we can see that another solution to the equation is θ=300^(∘). In fact, we see that, in the desired interval, the only solutions to the equation are θ = 60^(∘) and θ=300^(∘).

We will start by isolating cos^2 θ. Then, we will take square roots on both sides of the trigonometric equation to solve for cos θ.

Let's solve the obtained equations for θ between 180^(∘) and 360^(∘) one at a time.

Solving cos θ = 1/2

To solve the equation cos θ = 12, recall the graph of the cosine function. Let's also draw the line y = 12 to see for which values of θ we have cos θ = 12. Note that we only want solutions in the interval between 180^(∘) and 360^(∘).

We see that there is only one solution to our equation that falls in the interval between 180^(∘) and 360^(∘), which is θ = 300^(∘).

Solving cos θ = - 1/2

To solve the equation cos θ = - 12, let's use the graph of the cosine function again. This time we draw the line y = - 12 to see for which values of θ we have cos θ = - 12. Note that we only want solutions in the interval between 180^(∘) and 360^(∘).

We see that there is one solution to our equation in the desired interval, which is θ= 240^(∘).

Listing All Solutions

While solving the equations cos θ = 12 and cos θ = - 12 in the desired interval, we found the following solutions. 240^(∘) and 300^(∘) These are also all the solutions to our original equation in the desired interval.

We can check our answer by graphing y=4cos^2 θ and y=1 in the same coordinate plane on a graphic calculator. To do so, we must set the calculator in degree mode. We do this by pushing MODE and selecting Degree instead of Radian in the third row.

We want to find solutions between 180^(∘) and 360^(∘). Accordingly, let's resize the window of the calculator to show x-values between 180 and 360, and y-values between -4 and 4. To do so, we push WINDOW and change the settings.

Now, let's graph both functions. The x-coordinate of the points of intersection, if any, will be the solutions to the equation. Press the Y= button and type the functions in the first two rows. After writing the functions, push GRAPH to draw them.

We can see that there are two points of intersection. To find them, push 2nd and CALC and choose the fifth option, which is intersect.

Next, we need to choose left and right boundaries for one of the points. Finally, the calculator asks for a guess where the intersection point might be. After that, it will calculate the exact point for us. We have to do this two times, once for each point.

We see that all solutions match the ones we found previously.

We will start by rewriting the equation using the double angle identity for sine. sin 2 θ = 2 sin θ cos θ After that we will solve the obtained equation. Let's do these two things one at a time.

Rewriting the Equation

We will start by rewriting sin 2 θ as 2 sin θ cos θ. Let's do it! sin 2 θ - cos θ =0 ⇕ 2 cos θ sin θ - cos θ =0

Solving the Equation

We will now solve the obtained equation by factoring. 2 cos θ sin θ - cos θ =0 ⇕ cos θ ( 2 sin θ - 1 )= 0 Next, let's apply the Zero Product Property to solve for sin θ and cosθ.

Now, let's solve the obtained equations for θ in the desired interval.

Solving cos θ = 0

To solve the equation cos θ =0 in the desired interval, recall the graph of the cosine function, focusing on θ between 0 and π2. Let's also draw the line y = 0 to see for which values of θ we have cos θ = 0.

We see that the only solution to cos θ =0 in the desired interval is θ = π2.

Solving sin θ = 1/2

To solve the equation sin θ = 12 in the desired interval, recall the graph of the sine function, focusing on θ between 0 and π2. Let's also draw the line y = 12 to see for which values of θ we have sin θ = 12.

We see that the only solution to sin θ = 12 in the desired interval is θ = π6.

Listing All Solutions

While solving the equations cos θ =0 and sin θ = 12, we found the following solutions. π6 and π2 These are also all the solutions in the desired range to our original equation.

We can check our answer by graphing y=sin 2 θ-cos θ and y=0 in the same coordinate plane on a graphic calculator. We want to find solutions between 0 and π2 ≈ 1.57. Accordingly, let's resize the window of the calculator to show y-values between - 3 and 3, and x-values between 0 and 1.57. To do so, we push WINDOW and change the settings.

Now, let's graph both functions. The x-coordinate of the points of intersection, if any, will be the solutions to the equation. Press the Y= button and type the functions in the first two rows. After writing the functions, push GRAPH to draw them.

We can see that there are two points of intersection. To find them, push 2nd and CALC and choose the fifth option, which is intersect.

Next, we need to choose left and right boundaries for one of the points. Finally, the calculator asks for a guess where the intersection point might be. After that, it will calculate the exact point for us. We have to do this two times, once for each point.

Finally, to compare these results to the solutions we have previously found, let's use a calculator to approximate the latter ones.

As we can see, after approximating all the solutions we have previously found, they match the ones we got using the graphing calculator.

To solve the given equation, we will first factor the left-hand side of the equation and use the Zero Product Property to solve for sin θ. Then we will use the unit circle to find the exact values of θ that satisfy the equation.

Solving the Equation for sinθ

Let's start by factoring the expression on the left-hand side.

Next, we can use the Zero Product Property to solve the equation for sin θ .

Finding the Exact Values of θ

We obtained two values for sin θ . The sine of an angle in standard position is the y-coordinate of the point of intersection P of its terminal side and the unit circle. P(x,y)=(cosθ,sinθ) To solve the first equation, sin θ=1, we need to consider the points on the unit circle that have a y-coordinate of 1.

We found that θ= π2 is a solution for the equation sin θ =1. However, it does not belong to the interval π ≤ θ ≤ 2π, so we discard it. To solve the second equation, sin θ =- 1, we need to consider the points on the unit circle that have a y-coordinate of - 1.

We found one solution to the equation sin θ =- 1. This time, it does belong to the required interval. Therefore it is also a solution to the given equation. θ = 3π2 radians As a result, the given equation has only one solution that is between π and 2π.

To solve the given equation, we will first factor the equation and use the Zero Product Property to solve for sinθ. Then we will use the unit circle to find the exact values of θ that satisfy the equation.

Solving the Equation for sinθ

Let's start by factoring the expression on the left-hand side.

Next, we can use the Zero Product Property to solve the equation for sin θ.

Finding the Exact Values of θ

We obtained two values for sin θ. The sine of an angle in standard position is the y-coordinate of the point of intersection P of its terminal side and the unit circle. P(x,y)=(cosθ,sinθ) Let's start with the second equation. Since the unit circle has a radius of 1, no point that lies on it will ever have a y-coordinate of - 3. Therefore, we can disregard that equation. sin θ = - 3 doesnothave a solution To solve the equation sin θ=0, we need to consider the points on the unit circle that have a y-coordinate of 0.

Remember that a half turn measures π radians.

We found two solutions for the equation sin θ=0. Since we know that θ must be between π and 2π, we conclude that θ =π and θ = 2π are solutions to the given equation. θ=π radians and θ = 2π radians

Let's first use the definition of cosecant function and rewrite the given trigonometric equation.

To solve the trigonometric equation, let's use the Double Angle Identity for cosine. cos 2 θ = 1 - 2 sin^2 θ We can rewrite our equation using this identity. cos 2 θ sin θ = 1 ⇕ ( 1 - 2 sin^2 θ ) sin θ = 1 To solve the obtained equation we will first perform the substitution sin θ =t. This will simplify the notation. After solving the equation for t, we will go back to the original notation, and solve for θ.

Substituting sin θ = t

Let's substitute sin θ = t into the rewritten equation.

We want to solve the obtained polynomial equation. To do so, let P(t) be the polynomial on the left-hand side. We will identify the constant term and the leading coefficient. P(t) &= -2 t^3 +t -1 & ⇕ P(t) &= -2 t^3 + t +( - 1) Next, we will look for rational roots using the Rational Root Theorem. Any rational root will have the form ± p q, where p is a factor of the constant term - 1 and q is a factor of the leading coefficient -2. p: & ± 1 q: & ± 1, ± 2 Let's consider the possible combinations of ± p q until we find one that is a factor of the polynomial.

We found that t= -1 is a solution to the polynomial equation. By the Remainder Theorem, the remainder of dividing P(t) by (t - ( -1)) is zero. t-(- 1) ⇔ t+1 This means that (t+1) is a factor of P(t). Let's use synthetic division to factor P(t). To do this, we first need to rewrite P(t) so that all the coefficients are present. Any missing terms should be added to the polynomial with a coefficient of 0. P(t)=-2 t^3 + t -1 ⇕ P(t)=-2 t^3 + 0 t^2+t +(- 1) Now we can use synthetic division to factor out t+1.

Let's now rewrite P(t) using the factors we have found so far. P(t)=-2 t^3 + t -1 ⇕ P(t)=(t+1)( -2t^2+2t-1) We already know that t= -1 is a zero of our polynomial. To find the remaining zeros, we need to solve the quadratic equation -2t^2+2t-1 = 0. To do so, we will first identify a, b, and c. -2t^2+2t-1=0 ⇕ -2t^2+ 2t+( -1)=0 As we can see, a = -2, b = 2, and c = -1. Now, to solve the equation, we can use the Quadratic Formula.

We have a negative number under the radical sign. This means that there are no real solutions to the quadratic equation. Therefore, t= -1 is the only real root to the polynomial equation P(t)=0.

Solving for θ

Finally, let's return back from our substitution sin θ = t and solve the equation for θ. Recall that the only solution to the polynomial equation is t=- 1. t = -1 ⇔ sin θ = -1 As we can see, finding the solutions to our original equation is equivalent to finding θ such that sin θ = -1. Let's recall the graph of the sine function. We will also draw the line y = -1 to see for which values of θ we have sin θ = - 1.

We see that sin θ =- 1 at θ = 3π2 radians. We also see that the solutions appear every 2π. With this information, we can write the general form for the solutions to our equation. 3π/2 + 2kπ Here, k is any integer number.

Our first goal will be to rewrite the equation so that on one side we have only one trigonometric function, while on the other hand some constant.

Rewriting The Equation

To rewrite our equation, we will first divide both sides by sqrt(2).

Now, recall that sin π4 and cos π4 are both equal to sqrt(2)2. Therefore, in the equation above, we can replace the first sqrt(2)2 with cos π4 and the latter with sin π4.

Finally, we will use the following reverse Angle Sum Identity for sine. sin x cos y + cos x sin y = sin( x+ y) Using the above identity we are able to rewrite the left-hand side as a single trigonometric function. Let's do it! sin θ cos π/4 + cos θ sin π/4 = 1 ⇕ sin ( θ + π/4 ) = 1 Now that we have rewritten our equation, let's solve it!

Solving the Equation

To solve the equation sin ( θ + π4 ) = 1, let's first solve the equation sin x = 1. To do so, recall the graph of the sine function. Let's also draw the line y = 1 to see for which values of x we have sin x = 1.

We see that the first positive solution to the equation is x = π2. We also see that solutions repeat every 2π radians. With this information, we can write the general form for all the solutions to the equation sin x =1. x= π/2 + 2kπ Here, k is any integer number. In our case the input of sine function is θ + π4 instead of x. For this reason, we can substitute this expression for x, to get an equation for θ. x&= π/2 + 2kπ &⇕ θ + π/4 &= π/2+ 2kπ Finally, let's isolate θ in the obtained equation.