We are open Saturday and Sunday!

Call Now to Set Up Tutoring:

(888) 888-0446

Trigonometry : Quadratic Formula with Trigonometry

Study concepts, example questions & explanations for Trigonometry

Create An Account

All Trigonometry Resources

6 Diagnostic Tests 155 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #241 : Trigonometry

Which is not a solution for $\theta$ for $10 \sin ^2 \theta + \sin \theta - 3 = 0$ ?

Possible Answers:

150^o

143.13^o

60^o

323.13^o

30^o

Correct answer:

60^o

Explanation:

Using the quadratic formula gives:

$\sin \theta = \frac{ -1 \pm \sqrt{ 1 - 4(10)(-3)}}{2(10)} = \frac{-1 \pm 11}{20 } = \frac{1}{2} , -\frac{3}{5}$

$\theta = \sin ^{-1} (\frac{ 1}{2} ) = 30^o, 150^o$ or

$\theta = \sin ^{-1} (-\frac{3}{5}) \approx 323.13^o, 143.13^o$

Report an Error

Example Question #83 : Trigonometric Equations

Solve for $\theta$ : $0=2 \sin ^2 \theta - 3 \sin \theta + 1$

Possible Answers:

$\frac{\pi }{2} , \frac{3 \pi }{2} , \frac{\pi }{6} , \frac{5 \pi }{6}$

$\frac{\pi }{2} , \frac{\pi }{6} , \frac{5 \pi }{6}$

$\frac{\pi }{2} , \frac{\pi }{3}$

$\frac{\pi}{2} , \frac{\pi }{6} , \frac{11 \pi }{6}$

$\frac{\pi }{2} , \frac{\pi }{6}$

Correct answer:

$\frac{\pi }{2} , \frac{\pi }{6} , \frac{5 \pi }{6}$

Explanation:

Solve using the quadratic formula:

$\sin \theta = \frac{ 3 \pm \sqrt{9 - 4(2)}}{4(1)} = \frac{ 3 \pm \sqrt1}{4} = \frac{1}{2} \enspace or \enspace 1$

$\sin \theta = \frac{1}{2}$

$\theta = \sin ^{-1} (\frac{1}{2} ) = \frac{\pi }{6} , \frac{ 5 \pi }{6}$

$\sin \theta = 1$

$\theta = \sin ^{-1}(1)$

$\theta = \frac{\pi }{2}$

Report an Error

Example Question #21 : Quadratic Formula With Trigonometry

Find the roots for $y = \sin^2 x + 2 \sin x - 5$

Possible Answers:

60.56 ^o , 119.45^o

26.71^o, 153.29^o

No solution

90^o, 270^o

90^o

Correct answer:

No solution

Explanation:

To solve, use the quadratic formula:

$\sin \theta = \frac{ -2 \pm \sqrt{4 + 20 }}{2 } = \frac{-2 \pm \sqrt{24}}{2} = - \pm \sqrt{6}$

Both $-1 - \sqrt{6}$ and $-1+\sqrt{6}$ are outside of the range of the sine function, $-1 \leq \sin x \leq 1$ so there is no solution.

Report an Error

Example Question #22 : Quadratic Formula With Trigonometry

Solve for :

$0 = 2 \cos ^2 x + \cos x - 2$

Possible Answers:

38.67^o, 141.33^o

141.33^o, 321.33^o

16.31^o, 343.69^o

38.67^o, 321.33^o

38.67^o, 16.31^o

Correct answer:

38.67^o, 321.33^o

Explanation:

Solve using the quadratic formula:

$\cos x = \frac{-1 \pm \sqrt{1 + 16}}{4 } = \frac{-1 \pm \sqrt{17}}{4}$

$\frac{ -1 - \sqrt{17}}{4} \approx -1.28$ , outside the range for cosine.

$\frac{-1 + \sqrt{17}}{4} \approx 0.78$

$x = \cos ^{-1} (0.78 ) = 38.67^o$ according to a calculator.

The other angle with a cosine of 0.78 would be 360 ^o - 38.67^o = 321.33^o .

Report an Error

Example Question #23 : Quadratic Formula With Trigonometry

Solve for :

$0 = 4 \cos ^2 x - 21 \cos x + 5$

Possible Answers:

x = 75.52^o, 255.52 ^o

14.47^o, 345.52^o

x = 75.52^o, 104.48^o

x = 75.52^o, 284.48^o

14.47^o, 165.52^o

Correct answer:

x = 75.52^o, 284.48^o

Explanation:

Solve using the quadratic formula:

$\cos x = \frac{21 \pm \sqrt{441 - 4(4)(5)}}{2(4) } = \frac{ 21 \pm \sqrt{361}}{8} = \frac{21 \pm 19}{8}$

$\cos x = 5, \frac{1}{4}$

5 is outside the range for cosine, so the only solution that works is $\cos x = \frac{1}{4}$ :

$x = \cos ^{-1} (\frac{1}{4}) \approx 75.52 ^o$ according to a calculator

The other angle with a cosine of $\frac{1}{4}$ is 360^o - 75.52^o = 284.48^o

Report an Error

Example Question #24 : Quadratic Formula With Trigonometry

Solve for : $0 = 3 \cos ^2 x + 8 \cos x + 4$

Possible Answers:

48.19^o, 131.81^o

131.81^o, 228.19^o

48.19^o, 311.81^o

131.81^o, 311.81^o

48.19^o, 228.19^o

Correct answer:

131.81^o, 228.19^o

Explanation:

Use the quadratic formula:

$\cos x = \frac{ -8 \pm \sqrt{64 - 4(3)(4)}}{2(3)} = \frac{-8 \pm \sqrt{16}}{6 } = \frac{-8\pm4}{6}$

$\cos x = -2 \enspace or - \frac{2}{3}$

-2 is outside the range of cosine, so the answer has to come from $- \frac{2}{3}$ :

$\cos x = - \frac{2}{3}$

$x = \cos^{-1} (-\frac{2}{3} ) \approx 131.81^o$ according to a calculator

The other angle with a cosine of $-\frac{2}{3}$ is 360^o - 131.81^o = 228.19^o

Report an Error

Example Question #251 : Trigonometry

Solve the equation

2cos^2x+sinx=1

for $0\leq x\leq2\pi$ .

Possible Answers:

$\frac{\pi}{2},\frac{7\pi}{6}, \frac{11\pi}{6}$

$\frac{\pi}{2},\frac{\pi}{6}, \frac{2\pi}{3}$

$\frac{\pi}{2}$

$\frac{\pi}{2},\frac{\pi}{6}$

$\frac{7\pi}{6},\frac{11\pi}{6}$

Correct answer:

$\frac{\pi}{2},\frac{7\pi}{6}, \frac{11\pi}{6}$

Explanation:

First of all, we can use the Pythagorean identity sin^2x+cos^2x=1 to rewrite the given equation in terms of sinx .

2cos^2x+sinx=1

$\Rightarrow 2(1-sin^2x)+sinx=1$

$\Rightarrow 2-2sin^2x+sinx=1$

$\Rightarrow -2sin^2x+sinx+1=0$

$\Rightarrow 2sin^2x-sinx-1=0$

This is a quadratic equation in terms of sinx ; hence, we can use the quadratic formula to solve this equation for sinx .

$sinx=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

where a=2, b=-1, c=-1 .

$\Rightarrow sinx=\frac{1\pm \sqrt{(-1)^2-4(2)(-1)}}{2(2)}$

$\Rightarrow sinx=\frac{1\pm3}{4}$

$\Rightarrow sinx=-\frac{1}{2}, sinx=1$ .

Now, sinx=1 when $x=\frac{\pi}{2}$ , and $sinx=-\frac{1}{2}$ when $x=\frac{7\pi}{6}$ or $\frac{11\pi}{6}$ .

Hence, the solutions to the original equation 2cos^2x+sinx=1 are

$\frac{\pi}{2},\frac{7\pi}{6}, \frac{11\pi}{6}$

Report an Error

Example Question #26 : Quadratic Formula With Trigonometry

In the interval $0\leq x\leq\pi$ , what values of x satisfy the following equation?

$1-\sin x=\cos^2x-\frac{1}{4}$

Possible Answers:

$x=\frac{\pi}{3},\frac{2\pi}{3}$

$x=\frac{\pi}{4},\frac{7\pi}{12}$

$x=\frac{\pi}{6},\frac{5\pi}{6}$

$x=0,\pi$

Correct answer:

$x=\frac{\pi}{6},\frac{5\pi}{6}$

Explanation:

We start by rewriting the $\cos^2x$ term on the right hand side in terms of $\sin^2x$ .

$\cos^2x=1-\sin^2x$

$1-\sin x=1-\sin^2x-\frac{1}{4}$

We then move everything to the left hand side of the equation and cancel.

$1-\sin x-1+\sin^2x+\frac{1}{4}=0$

$\sin^2x-\sin x+\frac{1}{4}=0$

Apply the quadratic formula:

$\begin{align*} \sin x&=\frac{-(-1)\pm\sqrt{(-1)^2-4(1)(\frac{1}{4})}}{2(1)}\\ &=\frac{1\pm\sqrt{1-1}}{2}\\ &=\frac{1}{2} \end{align*}$

So $\sin x=\frac{1}{2}$ . Using the unit circle, the two values of that yield this are $x=\frac{\pi}{6}$ and $x=\frac{5\pi}{6}$ .

Report an Error

View Trigonometry Tutors

Dejuan
Certified Tutor

University of Illinois at Urbana-Champaign, Bachelor of Science, Electrical Engineering.

View Trigonometry Tutors

Leonard
Certified Tutor

Florida State University, Bachelor of Science, Environmental Science.

View Trigonometry Tutors

Yucheng
Certified Tutor

The University of Texas at Austin, Bachelor of Science, Mechanical Engineering.

All Trigonometry Resources

6 Diagnostic Tests 155 Practice Tests Question of the Day Flashcards Learn by Concept

Trigonometry Tutors in Top Cities:

Popular Courses & Classes

GRE Courses & Classes in Houston, Spanish Courses & Classes in Atlanta, Spanish Courses & Classes in Chicago, SAT Courses & Classes in Miami, GMAT Courses & Classes in San Francisco-Bay Area, SSAT Courses & Classes in Philadelphia, GMAT Courses & Classes in Los Angeles, SSAT Courses & Classes in Chicago, LSAT Courses & Classes in Boston, GRE Courses & Classes in San Francisco-Bay Area

Popular Test Prep

SSAT Test Prep in Miami, GRE Test Prep in Miami, ISEE Test Prep in Boston, GRE Test Prep in Denver, GMAT Test Prep in Phoenix, MCAT Test Prep in Boston, ACT Test Prep in San Francisco-Bay Area, LSAT Test Prep in Phoenix, GRE Test Prep in New York City, SAT Test Prep in Seattle

DMCA Complaint

If you believe that content available by means of the Website (as defined in our Terms of Service) infringes one or more of your copyrights, please notify us by providing a written notice (“Infringement Notice”) containing the information described below to the designated agent listed below. If Varsity Tutors takes action in response to an Infringement Notice, it will make a good faith attempt to contact the party that made such content available by means of the most recent email address, if any, provided by such party to Varsity Tutors.

Your Infringement Notice may be forwarded to the party that made the content available or to third parties such as ChillingEffects.org.

Please be advised that you will be liable for damages (including costs and attorneys’ fees) if you materially misrepresent that a product or activity is infringing your copyrights. Thus, if you are not sure content located on or linked-to by the Website infringes your copyright, you should consider first contacting an attorney.

Please follow these steps to file a notice:

You must include the following:

A physical or electronic signature of the copyright owner or a person authorized to act on their behalf; An identification of the copyright claimed to have been infringed; A description of the nature and exact location of the content that you claim to infringe your copyright, in \ sufficient detail to permit Varsity Tutors to find and positively identify that content; for example we require a link to the specific question (not just the name of the question) that contains the content and a description of which specific portion of the question – an image, a link, the text, etc – your complaint refers to; Your name, address, telephone number and email address; and A statement by you: (a) that you believe in good faith that the use of the content that you claim to infringe your copyright is not authorized by law, or by the copyright owner or such owner’s agent; (b) that all of the information contained in your Infringement Notice is accurate, and (c) under penalty of perjury, that you are either the copyright owner or a person authorized to act on their behalf.

Send your complaint to our designated agent at:

Charles Cohn Varsity Tutors LLC
101 S. Hanley Rd, Suite 300
St. Louis, MO 63105

Or fill out the form below:

Do not fill in this field

Contact Information

* Full legal name

Company name

* Phone number

* Email address

Address

* Address1

Address2

* City

* State

* Zipcode

* Country

Complaint Details

* URL of copyrighted work (where your original material is located)

* Please describe the copyrighted work so that it may be easily identified

I am the owner, or an agent authorized to act on behalf of the owner of an exclusive right that is allegedly infringed.

I have a good faith belief that the use of the material in the manner complained of is not authorized by the copyright owner, its agent, or the law.

This notification is accurate.

I acknowledge that there may be adverse legal consequences for making false or bad faith allegations of copyright infringement by using this process.

* Signature (your digital signature is legally binding)

HIGH SCHOOL

GRADUATE SCHOOL

K-8

Search 50+ Tests

math tutoring

science tutoring

foreign languages

elementary tutoring

other

Search 350+ Subjects

Trigonometry : Quadratic Formula with Trigonometry

Study concepts, example questions & explanations for Trigonometry

All Trigonometry Resources

Example Questions

Example Question #241 : Trigonometry

Example Question #83 : Trigonometric Equations

Example Question #21 : Quadratic Formula With Trigonometry

Example Question #22 : Quadratic Formula With Trigonometry

Example Question #23 : Quadratic Formula With Trigonometry

Example Question #24 : Quadratic Formula With Trigonometry

Example Question #251 : Trigonometry

Example Question #26 : Quadratic Formula With Trigonometry

All Trigonometry Resources

Report an issue with this question

DMCA Complaint

Contact Information

Address

Complaint Details

Email address:
Your name:
Feedback: