Question

Solve : 4-3sec²A = 0​

Answer 1

`\sf\:Negative = \boxed{\sf\:\theta =\pi n_(1)+(5\pi )/(6)\text{, }n_(1)\in \mathbb{Z}}\\\sf\: Positive = \boxed{\sf\:\theta =\pi n_(1)+(\pi )/(6)\text{, }n_(1)\in \mathbb{Z}}`

Explanation:

`4 - 3 \sec ^ { 2 } \theta = 0`

Let's solve this.

Isolate sec²θ.

`4 -3 \sec ^ { 2 } \theta = 0\\- 3 \sec^(2)\theta = -4\\3 \sec^(2)\theta=0\\\sec^(2) \theta=(4)/(3)`

Now, we now that, cos θ is the reciprocal of sec θ. Therefore,

`\sec^(2)\theta = (4)/(3)\\\cos^(2)\theta = (3)/(4)`

Bring the square root on both the sides of the equation to remove the square.

`\sqrt{\cos^(2)\theta} = \sqrt{(3)/(4) }\\\cos^(2)\theta = (+/-) (√(3) )/(2)`

If,

`\cos\theta = + (√(3))/(2)\\= cos \left( (\pi)/(6) \right)\\\Longrightarrow \boxed{\sf\:\theta =\pi n_(1)+(\pi )/(6)\text{, }n_(1)\in \mathbb{Z}}`

Also if,

`\cos\theta = -(√(3))/(2)\\= \cos \left(\pi - (\pi)/(6)\right)\\= \cos \left({(5\:\pi)/(6)\right)\\\Longrightarrow \boxed{\sf\:\theta =\pi n_(1)+(5\pi )/(6)\text{, }n_(1)\in \mathbb{Z}}`

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