Solve for all real values of r:1. 8 sin?(r - 3) +2 sin(3 – r) – 1 = 02. 2 sec(3.r + 5 27) + sec(673r5) = 7

Answered: Solve for all real values of r: 1. 8…

Calculus: Early Transcendentals

8th Edition

ISBN:9781285741550

Author:James Stewart

Publisher:James Stewart

Chapter1: Functions And Models

Section: Chapter Questions

Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...

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Question

Solve for all real values of r:
1. 8 sin?(r - 3) +2 sin(3 – r) – 1 = 0
2. 2 sec(3.r + 5 27) + sec(67
3r
5) = 7

Expert Solution

Step 1

“Since you have asked multiple question, we will solve the first question for you. If you want any specific question to be solved then please specify the question number or post only that question.”

Solution Question 1

Given,

$8 \sin^{2} (x - 3) + 2 \sin (3 - x) - 1 = 0$

We have to find the real values of $x$

Step 2

Solving

$8 \sin^{2} (x - 3) + 2 \sin (3 - x) - 1 = 0$

$\Rightarrow 8 \sin^{2} (x - 3) - 2 \sin (x - 3) - 1 = 0$ (because $\sin (- θ) = - \sin θ$ )

So, $8 \sin^{2} (x - 3) - 2 \sin (x - 3) - 1 = 0$ is a quadratic equation in $\sin (x - 3)$

Then, applying the quadratic formula

$\sin (x - 3) = \frac{- (- 2) \pm \sqrt{{(- 2)}^{2} - 4 \times 8 \times (- 1)}}{2 \times 8}$

$\sin (x - 3) = \frac{2 \pm \sqrt{4 + 32}}{16}$

$\sin (x - 3) = \frac{2 \pm \sqrt{36}}{16}$

$\sin (x - 3) = \frac{2 \pm 6}{16}$

So,

$\sin (x - 3) = \frac{2 + 6}{16} o r \sin (x - 3) = \frac{2 - 6}{16}$

$\sin (x - 3) = \frac{1}{2} o r \sin (x - 3) = - \frac{1}{4}$

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