Solve Trigonometric Equations and Inequalities in Quadratic Form - Pre-Calculus
Card 0 of 24
Solve for
in the equation
on the interval
.
Solve for in the equation
on the interval
.
If you substitute
you obtain a recognizable quadratic equation which can be solved for
,
.
Then we can plug
back into our equation and use the unit circle to find that
.
If you substitute you obtain a recognizable quadratic equation which can be solved for
,
.
Then we can plug back into our equation and use the unit circle to find that
.
Compare your answer with the correct one above
If
exists in the domain from
, solve the following: 
If exists in the domain from
, solve the following:
Factorize
.

Set both terms equal to zero and solve.



This value is not within the
domain.



This is the only correct value in the
domain.
Factorize .
Set both terms equal to zero and solve.
This value is not within the domain.
This is the only correct value in the domain.
Compare your answer with the correct one above
Given that theta exists from
, solve: 
Given that theta exists from , solve:
In order to solve
appropriately, do not divide
on both sides. The effect will eliminate one of the roots of this trig function.
Substract
from both sides.

Factor the left side of the equation.
![cos(\theta)[2cos(\theta)-1]=0](https://vt-vtwa-assets.varsitytutors.com/vt-vtwa/uploads/formula_image/image/363057/gif.latex)
Set each term equal to zero, and solve for theta with the restriction
.






The correct answer is:

In order to solve appropriately, do not divide
on both sides. The effect will eliminate one of the roots of this trig function.
Substract from both sides.
Factor the left side of the equation.
Set each term equal to zero, and solve for theta with the restriction .
The correct answer is:
Compare your answer with the correct one above
Solve
for 
Solve for
By subtracting
from both sides of the original equation, we get
. We know that the square of a trigonometric identity cannot be negative, regardless of the input, so there can be no solution.
By subtracting from both sides of the original equation, we get
. We know that the square of a trigonometric identity cannot be negative, regardless of the input, so there can be no solution.
Compare your answer with the correct one above
Solve
when 
Solve when
Given that, for any input,
, we know that
, and so the equation
can have no solutions.
Given that, for any input, , we know that
, and so the equation
can have no solutions.
Compare your answer with the correct one above
Solve
when 
Solve when
By adding one to both sides of the original equation, we get
, and by taking the square root of both sides of this, we get
From there, we get that, on the given interval, the only solutions are
and
.
By adding one to both sides of the original equation, we get , and by taking the square root of both sides of this, we get
From there, we get that, on the given interval, the only solutions are
and
.
Compare your answer with the correct one above
Solve for
in the equation
on the interval
.
Solve for in the equation
on the interval
.
If you substitute
you obtain a recognizable quadratic equation which can be solved for
,
.
Then we can plug
back into our equation and use the unit circle to find that
.
If you substitute you obtain a recognizable quadratic equation which can be solved for
,
.
Then we can plug back into our equation and use the unit circle to find that
.
Compare your answer with the correct one above
If
exists in the domain from
, solve the following: 
If exists in the domain from
, solve the following:
Factorize
.

Set both terms equal to zero and solve.



This value is not within the
domain.



This is the only correct value in the
domain.
Factorize .
Set both terms equal to zero and solve.
This value is not within the domain.
This is the only correct value in the domain.
Compare your answer with the correct one above
Given that theta exists from
, solve: 
Given that theta exists from , solve:
In order to solve
appropriately, do not divide
on both sides. The effect will eliminate one of the roots of this trig function.
Substract
from both sides.

Factor the left side of the equation.
![cos(\theta)[2cos(\theta)-1]=0](https://vt-vtwa-assets.varsitytutors.com/vt-vtwa/uploads/formula_image/image/363057/gif.latex)
Set each term equal to zero, and solve for theta with the restriction
.






The correct answer is:

In order to solve appropriately, do not divide
on both sides. The effect will eliminate one of the roots of this trig function.
Substract from both sides.
Factor the left side of the equation.
Set each term equal to zero, and solve for theta with the restriction .
The correct answer is:
Compare your answer with the correct one above
Solve
for 
Solve for
By subtracting
from both sides of the original equation, we get
. We know that the square of a trigonometric identity cannot be negative, regardless of the input, so there can be no solution.
By subtracting from both sides of the original equation, we get
. We know that the square of a trigonometric identity cannot be negative, regardless of the input, so there can be no solution.
Compare your answer with the correct one above
Solve
when 
Solve when
Given that, for any input,
, we know that
, and so the equation
can have no solutions.
Given that, for any input, , we know that
, and so the equation
can have no solutions.
Compare your answer with the correct one above
Solve
when 
Solve when
By adding one to both sides of the original equation, we get
, and by taking the square root of both sides of this, we get
From there, we get that, on the given interval, the only solutions are
and
.
By adding one to both sides of the original equation, we get , and by taking the square root of both sides of this, we get
From there, we get that, on the given interval, the only solutions are
and
.
Compare your answer with the correct one above
Solve for
in the equation
on the interval
.
Solve for in the equation
on the interval
.
If you substitute
you obtain a recognizable quadratic equation which can be solved for
,
.
Then we can plug
back into our equation and use the unit circle to find that
.
If you substitute you obtain a recognizable quadratic equation which can be solved for
,
.
Then we can plug back into our equation and use the unit circle to find that
.
Compare your answer with the correct one above
If
exists in the domain from
, solve the following: 
If exists in the domain from
, solve the following:
Factorize
.

Set both terms equal to zero and solve.



This value is not within the
domain.



This is the only correct value in the
domain.
Factorize .
Set both terms equal to zero and solve.
This value is not within the domain.
This is the only correct value in the domain.
Compare your answer with the correct one above
Given that theta exists from
, solve: 
Given that theta exists from , solve:
In order to solve
appropriately, do not divide
on both sides. The effect will eliminate one of the roots of this trig function.
Substract
from both sides.

Factor the left side of the equation.
![cos(\theta)[2cos(\theta)-1]=0](https://vt-vtwa-assets.varsitytutors.com/vt-vtwa/uploads/formula_image/image/363057/gif.latex)
Set each term equal to zero, and solve for theta with the restriction
.






The correct answer is:

In order to solve appropriately, do not divide
on both sides. The effect will eliminate one of the roots of this trig function.
Substract from both sides.
Factor the left side of the equation.
Set each term equal to zero, and solve for theta with the restriction .
The correct answer is:
Compare your answer with the correct one above
Solve
for 
Solve for
By subtracting
from both sides of the original equation, we get
. We know that the square of a trigonometric identity cannot be negative, regardless of the input, so there can be no solution.
By subtracting from both sides of the original equation, we get
. We know that the square of a trigonometric identity cannot be negative, regardless of the input, so there can be no solution.
Compare your answer with the correct one above
Solve
when 
Solve when
Given that, for any input,
, we know that
, and so the equation
can have no solutions.
Given that, for any input, , we know that
, and so the equation
can have no solutions.
Compare your answer with the correct one above
Solve
when 
Solve when
By adding one to both sides of the original equation, we get
, and by taking the square root of both sides of this, we get
From there, we get that, on the given interval, the only solutions are
and
.
By adding one to both sides of the original equation, we get , and by taking the square root of both sides of this, we get
From there, we get that, on the given interval, the only solutions are
and
.
Compare your answer with the correct one above
Solve for
in the equation
on the interval
.
Solve for in the equation
on the interval
.
If you substitute
you obtain a recognizable quadratic equation which can be solved for
,
.
Then we can plug
back into our equation and use the unit circle to find that
.
If you substitute you obtain a recognizable quadratic equation which can be solved for
,
.
Then we can plug back into our equation and use the unit circle to find that
.
Compare your answer with the correct one above
If
exists in the domain from
, solve the following: 
If exists in the domain from
, solve the following:
Factorize
.

Set both terms equal to zero and solve.



This value is not within the
domain.



This is the only correct value in the
domain.
Factorize .
Set both terms equal to zero and solve.
This value is not within the domain.
This is the only correct value in the domain.
Compare your answer with the correct one above