Solve each equation for the specified variable part a...

Solve each equation for the specified variable. Part A: 15y2−8y+1=0 for y. y= Part B: Solve the equation 15(sinθ)2−8sinθ+1=0 for θ. Find four solutions in the interval 0≤θ≤5. Give your answers in radians to three decimal places. Note: Use the result of Part A, substituting y=sinθ, to find two equations that can be solved for θ. Round your answers to three decimal places. θ= radians

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Transcript

00:01 So we need to use the quadratic formula to find all degree solutions and all solutions between 0 and 360 degrees.

00:07 And we're allowed to use the calculator to approximate our answers.

00:12 So remember that the quadratic formula is negative b plus or minus the square root of b squared minus 4ac all over 2a.

00:28 So we have the equation 2 sine squared of theta minus 2 sine of theta minus 1 equals 0.

00:47 So we're solving for sine of theta.

00:50 So sine of theta is equal to negative b plus or minus the square root of b squared minus 4ac, all over 2a.

00:59 So we have sine of theta is equal to negative negative 2.

01:08 Plus or minus the square root of negative 2 squared minus 4 times 2 times negative 1, the square root of all of that, all over 2 times a, which is 2, so all over 4.

01:34 This gives us sine of theta is equal to 2 plus or minus the square root of 12 over 4.

01:51 We plug this into a calculator, it's equal to 1 .366 and for 2 minus the square root of 12 over 4, this is negative 0 .366.

02:16 Okay, so we have that sine of theta for our first number is 1 .366, and sine of theta is equal to negative 0 .366.

02:32 So for this first portion, there are no solutions.

02:40 This is because the sign function pose from negative 1 to 1.

02:49 So there's no angle that will make sign of that angle 1 .366...