please solve the equations and show all the solutions between [0,2π). cos^2-3sin^2x=0  

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hala718 | High School Teacher | (Level 1) Educator Emeritus

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`cos^2 x - 3sin^2 x = 0`

We will use the trigonometric properties to solve.

We know that `cos^2 x + sin^2 x = 1`

`==gt cos^2 x = 1-sin^2 x`

==> Now we will substitute into the equation.

`==gt (1-sin^2 x) - 3sin^2 x = 0 `

`==gt 1- 4sin^2 x = 0 `

`==gt -4sin^2 x = -1 `

`==gt sin^2 x = 1/4 `

`\==gt sinx = +- 1/2`

`==gt x = pi/6, pi-pi/6, pi+pi/6, 2pi-pi/6`

==> x = { `pi/6, (5pi)/6, (7pi)/6, (11pi)/6` }

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