How do I solve for "x" in 3sec^2(x)+2+sin^2(x)-tan^2(x)+cos^2(x)=0?

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3sec^2(x)+2+sin^2(x)-tan^2(x)+cos^2(x)=0

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Answer 1 · 2018-02-08T22:32:16

There is no solution

If we group terms and recall our pythagorean identities, namely #sec^2x = tan^2x + 1# and #sin^2x +cos^2x = 1#, we can simplify to

#3sec^2x + 2 + 1 - tan^2x = 0#

If we solve for tangent in #sec^2x = tan^2x + 1#, we get #sec^2x - 1= tan^2x#, therefore:

#3sec^2x + 3 - (sec^2x - 1) = 0#

#3sec^2x + 3 - sec^2x + 1 = 0#

#2sec^2x + 4 = 0#

#sec^2x + 2 = 0#

#sec^2x = -2#

#x = O/#

Since the square root of a negative number is undefined there is no solution to the given equation.

As you can see in the following graph, there is no intersection between #y_1 = 3sec^2x + 2 + sin^2x - tan^2x + cos^2x # and #y_2 = 0#.

enter image source here

Hopefully this helps!