Solve for the first two positive solutions. 3 sin(x)+cos(x)=2

Name: Problem 40, Chapter 7: Precalculus: An Investigation of Functions
Uploaded: 2021-09-28T18:18:34-08:00
Duration: 5 min 14 s
Channel: Vijay Ahlawat
Description: Solve for the first two positive solutions. 3 sin(x)+cos(x)=2

Solve for the first two positive solutions. 3...

Key Concepts

Inverse Trigonometric Functions

These functions, such as arcsin, provide the means to determine the angle corresponding to a given trigonometric value. They are used to obtain the principal solution for equations like sin(?) = c, after which periodic properties are utilized to find additional solutions, which is essential when solving trigonometric equations.

Amplitude-Phase Transformation

This technique involves rewriting a linear combination of sine and cosine functions as a single sinusoidal function of the form R sin(x + ?) or R cos(x ? ?). This is achieved by calculating the amplitude R = ?(A² + B²) and a phase shift ? (or ?) based on the coefficients of the sine and cosine. It simplifies the solving process by reducing an equation with two trigonometric terms into one with a single trigonometric expression.

Trigonometric Equation Solving

This concept involves methods for finding all solutions to equations involving trigonometric functions. It typically includes rewriting the equation in a simpler form (often using the amplitude-phase transformation), applying inverse trigonometric functions to find the principal solution, and then using the periodicity of trigonometric functions to generate the general solutions.

Transcript

00:01 We have to solve the given equation start by getting rid of cos x from left side that is subtract cos x from both the sides we get 3 sine x equals 2 minus cos x and then a square on the both sides that is squaring the both sides we get 3 square is 9 sine square x equals this you can either foil or you can do a minus b whole square which will be 4 minus 2 times 2 that is 4 cos x plus cos square x now the sign square x we can replace this by 1 minus cos square x and keep the right side as it is cost square x now distribute this 9 minus 9 cos square x and shift all the terms to the left side that is subtract 4 add 4 cos x and subtract cost square x equals the right side will be 0 okay 9 minus 9 co square x minus cos square x so minus 10 cos square x then we have plus 4 cos x and 9 minus 4 is 5 equals equals zero now this or seems like a quadratic equation if we replace cos x let cause x equals t then we'll get minus 10 t square plus 4 t plus 5 equal to 0 we have a quadratic equation and we know how to solve the quadratic equation either we can solve it by factoring or if the factoring is not possible then we can also use the quadratic formula so this equation is a prime equation we cannot factor it so we'll use the quadratic formula that as t equals minus b plus minus square root b square minus 4 a c by 2a equals b is 4 so minus 4 plus minus square root of b square is 4 square is 16 minus 4 times minus 10 times 5 that is minus 200 and in the denominator we'll get two times minus 10 so this will be equal to minus 4 plus minus square root of 2 1 6 by minus 20 and this can be simplified to 2 plus minus 3 square root 6 by 10 these are the values of t that is t equals 2 plus 3 square root 6 by 10 or t equals 2 minus 3 square root 6 by 10 now t actually is cos x so it means cos x equals 2 plus 3 square root 6 by 10 or cause x equals 2 plus 3 square root 6 by 10 or cause x equals equals 2 minus 3 square root 6 by 10 now the question says we have to find the two first positive values that is we can directly solve for x that is x equals cos inverse 2 plus 3 square root 6 by 10 comma x equals cos inverse 2 minus 3 square root 6 by 10 and this we can solve using graphing calculator so x will be equal to 0 .363 radiance or 20 .8 degrees there also x equals 2 .135 radiance or 122 .33 degrees.

04:52 These are the approximate values as exact values cannot be determined...

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