I need to find all solutions of the given equations for the indicated interval. Round solutions to three decimal places if necessary.
1.) 3sin(x)+1=0, x within [0,2pi)
2.) 2sin(sq'd)(x)+cos(x)-1=0, x within R
3.) 4sin(sq'd)(x)-4sin(x)-1=0, x within R
4.) sin(x)+1=cos(x), x within [0, 2pi) -check for extraneous solutions.
on 2 remember that sin^2 x = 1-cos^2 x, then use the quadratic formula,or factor.
on 3, factor
on 4,try squaring both sides, then substitue to get rid of the cos^2
To find the solutions to these equations, we will use algebraic techniques and trigonometric identities. Let's go through each equation one by one.
1.) 3sin(x) + 1 = 0, x within [0,2pi)
To find the solutions within the interval [0,2pi), we need to find the values of x that satisfy the equation. Here's how you can proceed:
a. Subtract 1 from both sides of the equation: 3sin(x) = -1.
b. Divide both sides of the equation by 3: sin(x) = -1/3.
c. To find the solutions for sin(x) = -1/3, you can use inverse trigonometric functions. In this case, we will use arcsin or sin^(-1) function.
d. Calculate the inverse sine of -1/3 to find the solutions within the interval [0,2pi).
2.) 2sin^2(x) + cos(x) - 1 = 0, x within R
This equation involves both the sine and cosine functions. To solve it, follow these steps:
a. Rewrite sin^2(x) as (1 - cos^2(x)) using the identity sin^2(x) + cos^2(x) = 1.
b. Substitute (1 - cos^2(x)) for sin^2(x) in the equation: 2(1 - cos^2(x)) + cos(x) - 1 = 0.
c. Simplify the equation by distributing and combining like terms: 2 - 2cos^2(x) + cos(x) - 1 = 0.
d. Rearrange the equation to form a quadratic equation: -2cos^2(x) + cos(x) + 1 = 0.
e. Solve the quadratic equation using factoring, completing the square, or quadratic formula. Once you find the values of cos(x), substitute them back into the original equation to solve for x.
3.) 4sin^2(x) - 4sin(x) - 1 = 0, x within R
This is another quadratic equation involving the sine function. Here's what you can do to solve it:
a. Factor the quadratic equation, if possible. If the equation does not factor, you can use the quadratic formula to find the solutions. Once you find the solutions, substitute them back into the original equation to solve for x.
4.) sin(x) + 1 = cos(x), x within [0,2pi) - check for extraneous solutions
To solve this equation and check for extraneous solutions, follow these steps:
a. Rewrite cos(x) as sin(x + π/2) using the identity cos(x) = sin(x + π/2).
b. Substitute sin(x + π/2) for cos(x) in the equation: sin(x) + 1 = sin(x + π/2).
c. Simplify the equation by using the sum-to-product identity: sin(x) + 1 = sin(x)cos(π/2) + cos(x)sin(π/2).
d. Rearrange the equation to form a quadratic equation: sin(x) - sin(x)cos(π/2) - cos(x)sin(π/2) + 1 = 0.
e. Simplify further: sin(x)(1 - cos(π/2)) - cos(x)(sin(π/2)) + 1 = 0.
f. Apply the values of cos(π/2) and sin(π/2) to the equation and simplify.
g. Solve the resulting equation and check for extraneous solutions by substituting them back into the original equation to verify if they satisfy the equation.
Remember to round the solutions to three decimal places, if necessary, based on the given instructions.