Solve for x: tan^2 x - square root 3=0 and  2cos^2 x - 3sinx -3 =0

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

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(a) `tan^2(x)-sqrt(3)=0 `

`tan^2(x)=sqrt(3) `

`tan(x)=pm sqrt(sqrt(3)) `

`x=tan^(-1)sqrt(sqrt(3)) `

`x~~.9210 +npi ` or `x~~2.2206+npi ` in radians, where n is an integer.

** Was this supposed to be `tan^2(x)-3=0 ` ? If so, `tan(x)=pm sqrt(3) ` and `x=pi/3+npi ` or `x=(2pi)/3+npi ` .**

(b) `2cos^2(x)-3sin(x)-3=0 `

This looks almost like a quadratic in sin(x), so we rewrite the term with cosine squared in terms of sine using the pythagorean relationship.

`cos^2(x)+sin^2(x)=1 ==> cos^2(x)=1-sin^2(x) ` so:

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

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

`2sin^2(x)+3sin(x)+1=0 `

`(2sin(x)+1)(sin(x)+1)=0 `

`sin(x)=-1/2 `  or sin(x)=-1

`x=sin^(-1)(-1/2) ` so `x=(11pi)/6+2npi ` or `x=(7pi)/6+2npi ` where n is an integer or `x=(3pi)/2+2npi ` .

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