### All Precalculus Resources

## Example Questions

### Example Question #1 : Solve Trigonometric Equations And Inequalities In Quadratic Form

If exists in the domain from , solve the following:

**Possible Answers:**

**Correct answer:**

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.

### Example Question #1 : Solving Trigonometric Equations And Inequalities

Solve for in the equation on the interval .

**Possible Answers:**

**Correct answer:**

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

.

### Example Question #1 : Solving Trigonometric Equations And Inequalities

Given that theta exists from , solve:

**Possible Answers:**

**Correct answer:**

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:

### Example Question #4 : Solve Trigonometric Equations And Inequalities In Quadratic Form

Solve for

**Possible Answers:**

There is no solution.

**Correct answer:**

There is 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.

### Example Question #5 : Solve Trigonometric Equations And Inequalities In Quadratic Form

Solve when

**Possible Answers:**

There are no solutions.

**Correct answer:**

There are no solutions.

Given that, for any input, , we know that, and so the equation can have no solutions.

### Example Question #1 : Solve Trigonometric Equations And Inequalities In Quadratic Form

Solve when

**Possible Answers:**

There are no solutions.

**Correct answer:**

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 .

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