Hey just came across this 6 mark question , help please?? (b) Solve, for 0 ≤ θ < 360°, the equation 2 tan2 θ + sec θ = 1, giving your answers to 1 decimal place.
is that tangent squared?
Using sin^2x+cos^2x = 1 identity seems like a good start?
\[\sec \theta = 1 - 2\tan2\theta\] square both sides write sec^2 as 1 + tan^2 we get 4[\tan2\theta ^{2} = \tan ^{2}\] sq root it 2 tan2(theta)=tantheta
2sin^2 x + cosx = cos^2x This should result in a quadratic equation of cos x. Then, solve?
tan2theta = (2tan theta)/( 1+ tan^2)
hmmmm...that seems convoluted.
short way .. follow gt ..
sorry messed up in the equation writer basically i converted the equation into tan(theta)
I messed up as well. good job GT
im getting theta = 0 or pi/3
but hey kinzan is that squared or 2theta ??
because i have solved for 2 theta and gt has solved taking squared
hey guys yeah it is tan^2
then follow gt 's answer
I did not give the answer. Rather the approach to convert the equation into a quadratic equation of cos(theta), which when solved will lead to the answer.
Add two to both sides first.
\[2\tan ^{2}\theta+2+\sec \theta=\]=3
Now factor 2 out of the first two terms:
\[2(\tan ^{2}\theta+1)+\sec \theta=3\]
Now substitute sec^2 for tan^2 + 1 and subtract 3 from both sides.
\[2\sec ^{2}\theta+\sec \theta-3=0\]
Now factor:
\[(2\sec \theta+3)(\sec \theta-1)=0\]
Can you get it from here?
\[\sec \theta=-1.5 or \sec \theta=1\]
theta = 2.3 radians or 0 radians
@Mertsj: Thanks ! But how do you work out theta= 2.3 radians on your calculator? Do you use the rule : sec θ=1/cosθ ??
Yes
So cos theta = -2/3 or 1
you mean -3/2 right??
No. sec theta = -3/2 so cos theta = -2/3
So type -.666666666666666 into your calculator (which is set in radian mode) and hit the inverse cosine button.
Or if you calculator does fractions, type in -2/3 and push inverse cosine
i got 2.3?
theta = 2.3 radians or 0 radians
ohh sorry in degrees mode i got 131.8 which is correct!
yes, thanks! :D
Both are correct. 131.8 degrees or 2.3 radians. don't forget the 0 degrees or 0 radiana
thanks :)
yw
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