Solve on the interval 0 2pi

👉 Learn how to solve trigonometric equations. There are various methods that can be used to evaluate trigonometric equations, they include factoring out the

Solve on the interval [0,2pi)1+ sin 0=1/2​

i!The solutions are x=pi/3,pi,(5pi)/3 Use this identity: cos2x=2cos^2x-1 Now, here's the actual problem: 2cos2x+2cosx=0 2(color(red)(cos2x))+2cosx=0 2(color(red)(2cos^2x-1))+2cosx=0 color(red)(4cos^2x-2)+2cosx=0 4cos^2x+2cosx-2=0 4(cosx)^2+2cosx-2=0 Substitute u for cosx: 4u^2+2u-2=0 2u^2+u-1=0 (2u-1)(u+1)=0 u=1/2,-1 Put cosx back in for u: cosx=1/2, qquadcosx=-1 x=pi/3,(5pi)/3, pi Hope this helped!

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solve on the interval 0 2pi 4 sec x+6=-2

sin(2x) = 1 2 sin ( 2 x) = 1 2. Take the inverse sine of both sides of the equation to extract x x from inside the sine. 2x = arcsin(1 2) 2 x = arcsin ( 1 2) Simplify the right side. Tap for more steps
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Solve on the interval [0,2pi) tanx=2.3

👉 Learn how to solve trigonometric equations. There are various methods that can be used to evaluate trigonometric identities, they include by factoring out

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Solve the equation over the interval [0,2pi)?

Solve the equation on the interval [0, 2 π). Write numbers as integers of simplifed fractions and separate multiple answers with a comma. Write numbers as integers of simplifed fractions and

How do I solve for x on the interval [0, 2pi]

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